Continuity and Differentiabilty: Chapter 5 Links
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Second Order Derivative 5.7 deals with finding the derivative of a derivative. If y=f(x), then its first derivative is \frac{dy}{dx}. Differentiating \frac{dy}{dx} once again with respect to x gives the second order derivative, denoted by \frac{d^2y}{dx^2}.
\displaystyle \frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)
The notation \displaystyle \frac{d^2y}{dx^2} is read as “dee two y by dee x square”. It denotes the derivative of the first derivative \frac{dy}{dx} with respect to x.
In this exercise, we will learn how to find second order derivatives of various functions using the rules of differentiation and apply the concept to solve NCERT questions.
Key Concepts
1. First Derivative
If y=f(x), then the derivative of y with respect to x is called the first derivative of y.
\displaystyle \frac{dy}{dx}=f'(x)
The first derivative represents the rate of change of y with respect to x.
Example: If y=x^3, then \displaystyle \frac{dy}{dx}=3x^2.
2. Second Order Derivative
The derivative of the first derivative is called the second order derivative or second derivative.
\displaystyle \frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)
It measures the rate of change of the first derivative.
Example: If y=x^3, then
\displaystyle \frac{dy}{dx}=3x^2
and
\displaystyle \frac{d^2y}{dx^2}=6x
3. Notations for the Second Derivative
The second derivative can be represented in different notations.
- \displaystyle \frac{d^2y}{dx^2}
- y''
- f''(x)
- y_2
All these symbols denote the second derivative of a function.
4. Finding Second Derivatives
- Find the first derivative \displaystyle \frac{dy}{dx}.
- Differentiate \displaystyle \frac{dy}{dx} once again with respect to x.
- Simplify the resulting expression to obtain \displaystyle \frac{d^2y}{dx^2}.
Example: If y=\sin x, then
\displaystyle \frac{dy}{dx}=\cos x
and
\displaystyle \frac{d^2y}{dx^2}=-\sin x
5. Common Rules Used
While finding second derivatives, the same differentiation rules used for first derivatives continue to apply.
- Power Rule
- Chain Rule
- Product Rule
- Quotient Rule
- Derivatives of trigonometric, inverse trigonometric, exponential and logarithmic functions
Tip: While solving questions on Second Order Derivative, always find \frac{dy}{dx} carefully and simplify it before proceeding further. Since \frac{d^2y}{dx^2} is obtained by differentiating \frac{dy}{dx}, any mistake in the first derivative will be carried forward to the final answer.
Let us now solve all the NCERT questions step by step in this exercise 5.7 of Continuity and Differentiability.
Question 1: Second Order Derivative 5.7
1. Find \displaystyle \frac{d^2y}{dx^2} if y=x^2+3x+2.
Solution
Given,
y=x^2+3x+2
Differentiating w.r.t. x, we get
\displaystyle \frac{dy}{dx}=\frac{d}{dx}(x^2+3x+2)
\displaystyle =2x+3 \qquad ...(1)
Differentiating (1) w.r.t. x, we get
\displaystyle \frac{d^2y}{dx^2}=\frac{d}{dx}(2x+3)
\displaystyle =2
Hence,
\boxed{\displaystyle \frac{d^2y}{dx^2}=2}
Question 2: Second Order Derivatives 5.7
2. Find \displaystyle \frac{d^2y}{dx^2} if y=x^{20}.
Solution
Given,
y=x^{20}
Differentiating w.r.t. x, we get
\displaystyle \frac{dy}{dx}=\frac{d}{dx}(x^{20})
\displaystyle =20x^{19} \qquad ...(1)
Differentiating (1) w.r.t. x,
We get, \displaystyle \frac{d^2y}{dx^2}=\frac{d}{dx}(20x^{19})
\displaystyle =20\times19\,x^{18}
\displaystyle =380x^{18}
Hence,
\boxed{\displaystyle \frac{d^2y}{dx^2}=380x^{18}}
You may already be following Maths Better for NCERT Solutions for the topics like
Likewise this Exercise 5.7 of Continuity and Differentiability for Class 12 Maths is designed to strengthen your concepts and improve step-by-step problem-solving skills. Now, let’s move on to the next question.
Question 3: Continuity and Differentiability 5.7
3. Find \displaystyle \frac{d^2y}{dx^2} if y=x\cos x.
Solution
Given,
y=x\cos x
Differentiating w.r.t. x using the Product Rule,
We get, \displaystyle \frac{dy}{dx}=x\frac{d}{dx}(\cos x)+\cos x\frac{d}{dx}(x)
\displaystyle =x(-\sin x)+\cos x(1)
\displaystyle =\cos x-x\sin x \qquad ...(1)
Differentiating (1) w.r.t. x, we get
\displaystyle \frac{d^2y}{dx^2}=\frac{d}{dx}(\cos x)-\frac{d}{dx}(x\sin x)
Using the Product Rule,
\displaystyle \frac{d}{dx}(x\sin x)=x\cos x+\sin x
Therefore,
\displaystyle \frac{d^2y}{dx^2}=-\sin x-(x\cos x+\sin x)
\displaystyle =-x\cos x-2\sin x
Hence,
\boxed{\displaystyle \frac{d^2y}{dx^2}=-x\cos x-2\sin x}
Question 4: Second Order Derivative
4. Find \displaystyle \frac{d^2y}{dx^2} if y=\log x.
Solution
Given,
y=\log x
Differentiating w.r.t. x, we get
\displaystyle \frac{dy}{dx}=\frac{d}{dx}(\log x)
\displaystyle =\frac{1}{x} \qquad ...(1)
Differentiating (1) w.r.t. x,
We get, \displaystyle \frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{1}{x}\right)
\displaystyle =\frac{d}{dx}(x^{-1})
\displaystyle =-x^{-2}
or
\displaystyle =-\frac{1}{x^2}
Hence,
\boxed{\displaystyle \frac{d^2y}{dx^2}=-\frac{1}{x^2}}
Question 5: Second Order Derivative 5.7
5. Find \displaystyle \frac{d^2y}{dx^2} if y=x^3\log x.
Solution
Given,
y=x^3\log x
Differentiating w.r.t. x using the Product Rule,
We get, \displaystyle \frac{dy}{dx}=x^3\frac{d}{dx}(\log x)+\log x\frac{d}{dx}(x^3)
\displaystyle =x^3\left(\frac{1}{x}\right)+\log x(3x^2)
\displaystyle =x^2+3x^2\log x
i.e. \displaystyle \frac{dy}{dx}=x^2(1+3\log x) \qquad ...(1)
Differentiating (1) w.r.t. x using the Product Rule,
We get, \displaystyle \frac{d^2y}{dx^2}=x^2\frac{d}{dx}(1+3\log x)+(1+3\log x)\frac{d}{dx}(x^2)
\displaystyle =x^2\left(\frac{3}{x}\right)+(1+3\log x)(2x)
\displaystyle =3x+2x+6x\log x
or
\displaystyle =5x+6x\log x
\displaystyle =x(5+6\log x)
Hence,
\boxed{\displaystyle \frac{d^2y}{dx^2}=x(5+6\log x)}
Important: If y=f(x), then the derivative \frac{dy}{dx} is called the first derivative of y. Differentiating \frac{dy}{dx} once again with respect to x gives the second order derivative \frac{d^2y}{dx^2}.
Question 6: Higher Order Derivative 5.7
6. Find \displaystyle \frac{d^2y}{dx^2} if y=e^x\sin 5x.
Solution
Given,
y=e^x\sin 5x
Differentiating w.r.t. x using the Product Rule,
We get, \displaystyle \frac{dy}{dx}=e^x\frac{d}{dx}(\sin 5x)+\sin 5x\frac{d}{dx}(e^x)
\displaystyle =e^x(5\cos 5x)+e^x\sin 5x
\displaystyle =e^x(5\cos 5x+\sin 5x) \qquad ...(1)
Differentiating (1) w.r.t. x using the Product Rule,
We get, \displaystyle \frac{d^2y}{dx^2}=e^x\frac{d}{dx}(5\cos 5x+\sin 5x)+(5\cos 5x+\sin 5x)\frac{d}{dx}(e^x)
\displaystyle =e^x(-25\sin 5x+5\cos 5x)+e^x(5\cos 5x+\sin 5x)
\displaystyle =e^x(-24\sin 5x+10\cos 5x)
or
\displaystyle =2e^x(5\cos 5x-12\sin 5x)
Hence,
\boxed{\displaystyle \frac{d^2y}{dx^2}=2e^x(5\cos 5x-12\sin 5x)}
Question 7: Second Order Derivative 5.7
7. Find \displaystyle \frac{d^2y}{dx^2} if y=e^{6x}\cos 3x.
Solution
Given,
y=e^{6x}\cos 3x
Differentiating w.r.t. x using the Product Rule,
We get, \displaystyle \frac{dy}{dx}=e^{6x}\frac{d}{dx}(\cos 3x)+\cos 3x\frac{d}{dx}(e^{6x})
\displaystyle =e^{6x}(-3\sin 3x)+6e^{6x}\cos 3x
\displaystyle =e^{6x}(6\cos 3x-3\sin 3x) \qquad ...(1)
Differentiating (1) w.r.t. x using the Product Rule,
We get, \displaystyle \frac{d^2y}{dx^2}=e^{6x}\frac{d}{dx}(6\cos 3x-3\sin 3x)+(6\cos 3x-3\sin 3x)\frac{d}{dx}(e^{6x})
\displaystyle =e^{6x}(-18\sin 3x-9\cos 3x)+6e^{6x}(6\cos 3x-3\sin 3x)
or
\displaystyle =e^{6x}(-18\sin 3x-9\cos 3x+36\cos 3x-18\sin 3x)
\displaystyle =e^{6x}(27\cos 3x-36\sin 3x)
or
\displaystyle =9e^{6x}(3\cos 3x-4\sin 3x)
Hence,
\boxed{\displaystyle \frac{d^2y}{dx^2}=9e^{6x}(3\cos 3x-4\sin 3x)}
Question 8: Second Order Derivative 5.7
8. Find \displaystyle \frac{d^2y}{dx^2} if y=\tan^{-1}x.
Solution
Given,
y=\tan^{-1}x
Differentiating w.r.t. x, we get
\displaystyle \frac{dy}{dx}=\frac{d}{dx}\left(\tan^{-1}x\right)
\displaystyle =\frac{1}{1+x^2} \qquad ...(1)
Differentiating (1) w.r.t. x,
We get, \displaystyle \frac{d^2y}{dx^2}=\frac{d}{dx}(1+x^2)^{-1}
\displaystyle =-(1+x^2)^{-2}\cdot\frac{d}{dx}(1+x^2)
\displaystyle =-(1+x^2)^{-2}(2x)
or
\displaystyle =-\frac{2x}{(1+x^2)^2}
Hence,
\boxed{\displaystyle \frac{d^2y}{dx^2}=-\frac{2x}{(1+x^2)^2}}
NCERT Class 12 Maths includes several important exercises across both Part 1 and Part 2, and I’ll cover them one by one with clear explanations and step-by-step solutions. Many of these questions are also available in video format on my YouTube channel, @MathsBetter, to help you understand the concepts more visually.
Now, let’s proceed to the next question.
Question 9: Derivative of Second Order
9. Find \displaystyle \frac{d^2y}{dx^2} if y=\log(\log x).
Solution
Given,
y=\log(\log x)
Differentiating w.r.t. x using the Chain Rule,
We get, \displaystyle \frac{dy}{dx}=\frac{1}{\log x}\cdot\frac{d}{dx}(\log x)
\displaystyle =\frac{1}{\log x}\cdot\frac{1}{x}
\displaystyle =\frac{1}{x\log x} \qquad ...(1)
Differentiating (1) w.r.t. x, we get
\displaystyle \frac{d^2y}{dx^2}=\frac{d}{dx}(x\log x)^{-1}
\displaystyle =-(x\log x)^{-2}\cdot\frac{d}{dx}(x\log x)
Using the Product Rule,
\displaystyle \frac{d}{dx}(x\log x)=x\cdot\frac{1}{x}+\log x\cdot1
\displaystyle =1+\log x
Therefore,
\displaystyle \frac{d^2y}{dx^2}=-(x\log x)^{-2}(1+\log x)
\displaystyle =-\frac{1+\log x}{x^2(\log x)^2}
Hence,
\boxed{\displaystyle \frac{d^2y}{dx^2}=-\frac{1+\log x}{x^2(\log x)^2}}
Question 10: Second Order Derivative 5.7
10. Find \displaystyle \frac{d^2y}{dx^2} if y=\sin(\log x).
Solution
Given,
y=\sin(\log x)
Differentiating w.r.t. x using the Chain Rule, we get
\displaystyle \frac{dy}{dx}=\cos(\log x)\cdot\frac{d}{dx}(\log x)
\displaystyle =\frac{\cos(\log x)}{x} \qquad ...(1)
Differentiating (1) w.r.t. x using the Quotient Rule, we get
\displaystyle \frac{d^2y}{dx^2}=\frac{x\frac{d}{dx}\big(\cos(\log x)\big)-\cos(\log x)\frac{d}{dx}(x)}{x^2}
Now,
\displaystyle \frac{d}{dx}\big(\cos(\log x)\big)=-\sin(\log x)\cdot\frac{1}{x}
Therefore,
\displaystyle \frac{d^2y}{dx^2}=\frac{x\left(-\frac{\sin(\log x)}{x}\right)-\cos(\log x)}{x^2}
\displaystyle =-\frac{\sin(\log x)+\cos(\log x)}{x^2}
Hence,
\boxed{\displaystyle \frac{d^2y}{dx^2}=-\frac{\sin(\log x)+\cos(\log x)}{x^2}}
Proof-based questions involving y, y_1 and y_2 can often be solved more efficiently by manipulating the first derivative before finding the second derivative.
Question 11: Continuity and Differentiability 5.7
11. If y=5\cos x-3\sin x, prove that \displaystyle \frac{d^2y}{dx^2}+y=0.
Solution
Given,
y=5\cos x-3\sin x
Differentiating w.r.t. x, we get
\displaystyle \frac{dy}{dx}=5(-\sin x)-3(\cos x)
\displaystyle =-5\sin x-3\cos x \qquad ...(1)
Differentiating (1) w.r.t. x, we get
\displaystyle \frac{d^2y}{dx^2}=-5\cos x+3\sin x
\displaystyle =-(5\cos x-3\sin x)
Since y=5\cos x-3\sin x,
\displaystyle \frac{d^2y}{dx^2}=-y
Therefore,
\displaystyle \frac{d^2y}{dx^2}+y=-y+y
\displaystyle =0
Hence,
\boxed{\displaystyle \frac{d^2y}{dx^2}+y=0}
In some questions, the second derivative is required in terms of y alone. In such cases, it is useful to rewrite the given relation and substitute appropriately after differentiation.
Question 12: Second Order Derivative 5.7
12. If y=\cos^{-1}x, find \displaystyle \frac{d^2y}{dx^2} in terms of y alone.
Solution
Given,
y=\cos^{-1}x
Differentiating w.r.t. x, we get
\displaystyle \frac{dy}{dx}=-\frac{1}{\sqrt{1-x^2}} \qquad ...(1)
Differentiating (1) w.r.t. x,
We get, \displaystyle \frac{d^2y}{dx^2}=-\frac{d}{dx}(1-x^2)^{-1/2}
\displaystyle =-\left(-\frac{1}{2}\right)(1-x^2)^{-3/2}(-2x)
\displaystyle =-\frac{x}{(1-x^2)^{3/2}} \qquad ...(2)
Since y=\cos^{-1}x, we have
x=\cos y
and
\displaystyle 1-x^2=1-\cos^2y=\sin^2y
Substituting these values in (2),
We get, \displaystyle \frac{d^2y}{dx^2}=-\frac{\cos y}{(\sin^2y)^{3/2}}
\displaystyle =-\frac{\cos y}{\sin^3y}
\displaystyle =-\cot y\,\csc^2y
Hence,
\boxed{\displaystyle \frac{d^2y}{dx^2}=-\cot y\,\csc^2y}
Question 13: Second Order Derivative 5.7
13. If y=3\cos(\log x)+4\sin(\log x), show that x^2y_2+xy_1+y=0.
Solution
Given,
y=3\cos(\log x)+4\sin(\log x)
Differentiating w.r.t. x, we get
\displaystyle y_1=\frac{dy}{dx}=3\left[-\sin(\log x)\cdot\frac{1}{x}\right]+4\left[\cos(\log x)\cdot\frac{1}{x}\right]
\displaystyle =\frac{-3\sin(\log x)+4\cos(\log x)}{x} \qquad ...(1)
Differentiating (1) w.r.t. x using the Quotient Rule, we get
\displaystyle y_2=\frac{x(-3\sin(\log x)+4\cos(\log x))'-(-3\sin(\log x)+4\cos(\log x))}{x^2}
Therefore,
\displaystyle y_2=\frac{-3\cos(\log x)-4\sin(\log x)+3\sin(\log x)-4\cos(\log x)}{x^2}
\displaystyle =\frac{3\sin(\log x)-7\cos(\log x)-4\sin(\log x)}{x^2}
i.e. \displaystyle y_2=-\frac{\sin(\log x)+7\cos(\log x)}{x^2} \qquad ...(2)
Now,
\displaystyle x^2y_2=-(\sin(\log x)+7\cos(\log x))
and
\displaystyle xy_1=-3\sin(\log x)+4\cos(\log x)
Adding x^2y_2, xy_1 and y,
We get, \displaystyle x^2y_2+xy_1+y
\displaystyle =[-\sin(\log x)-7\cos(\log x)]+[-3\sin(\log x)+4\cos(\log x)]+[3\cos(\log x)+4\sin(\log x)]
\displaystyle =0
Hence,
\boxed{\displaystyle x^2y_2+xy_1+y=0}
Question 14: Derivative of Second Order
14. If y=Ae^{mx}+Be^{nx}, show that \displaystyle \frac{d^2y}{dx^2}-(m+n)\frac{dy}{dx}+mny=0.
Solution
Given,
y=Ae^{mx}+Be^{nx}
Differentiating w.r.t. x, we get
\displaystyle \frac{dy}{dx}=Am\,e^{mx}+Bn\,e^{nx} \qquad ...(1)
Differentiating (1) w.r.t. x, we get
\displaystyle \frac{d^2y}{dx^2}=Am^2e^{mx}+Bn^2e^{nx} \qquad ...(2)
Now, \displaystyle \frac{d^2y}{dx^2}-(m+n)\frac{dy}{dx}+mny
\displaystyle =(Am^2e^{mx}+Bn^2e^{nx})-(m+n)(Ame^{mx}+Bne^{nx})+mn(Ae^{mx}+Be^{nx})
\displaystyle =Ae^{mx}\big[m^2-m(m+n)+mn\big]+Be^{nx}\big[n^2-n(m+n)+mn\big]
(Simplifying the brackets)
\displaystyle =Ae^{mx}(m^2-m^2-mn+mn)+Be^{nx}(n^2-mn-n^2+mn)
\displaystyle =Ae^{mx}(0)+Be^{nx}(0)
i.e.
\displaystyle =0
Hence,
\boxed{\displaystyle \frac{d^2y}{dx^2}-(m+n)\frac{dy}{dx}+mny=0}
Question 15: Second Order Derivatives 5.7
15. If y=500e^{7x}+600e^{-7x}, show that \displaystyle \frac{d^2y}{dx^2}=49y.
Solution
Given,
y=500e^{7x}+600e^{-7x}
Differentiating w.r.t. x, we get
\displaystyle \frac{dy}{dx}=500(7e^{7x})+600(-7e^{-7x})
\displaystyle =3500e^{7x}-4200e^{-7x} \qquad ...(1)
Differentiating (1) w.r.t. x,
We get, \displaystyle \frac{d^2y}{dx^2}=3500(7e^{7x})-4200(-7e^{-7x})
\displaystyle =24500e^{7x}+29400e^{-7x}
\displaystyle =49(500e^{7x}+600e^{-7x})
Since y=500e^{7x}+600e^{-7x},
\displaystyle \frac{d^2y}{dx^2}=49y
Hence,
\boxed{\displaystyle \frac{d^2y}{dx^2}=49y}
Important: While finding second order derivatives, all the standard rules of differentiation such as the Product Rule, Quotient Rule and Chain Rule continue to apply. Therefore, a strong understanding of first derivatives is essential.
Question 16: Second Order Derivative 5.7
16. If e^y(x+1)=1, show that \displaystyle \frac{d^2y}{dx^2}=\left(\frac{dy}{dx}\right)^2.
Solution
Given,
e^y(x+1)=1
Differentiating both sides w.r.t. x, we get
\displaystyle \frac{d}{dx}\big(e^y(x+1)\big)=\frac{d}{dx}(1)
Using the Product Rule,
\displaystyle e^y(x+1)\frac{dy}{dx}+e^y=0
\displaystyle e^y\left((x+1)\frac{dy}{dx}+1\right)=0
Since e^y\neq0,
\displaystyle (x+1)\frac{dy}{dx}+1=0
\displaystyle \frac{dy}{dx}=-\frac{1}{x+1} \qquad ...(1)
Differentiating (1) w.r.t. x,
We get, \displaystyle \frac{d^2y}{dx^2}=\frac{d}{dx}\left(-(x+1)^{-1}\right)
\displaystyle =(x+1)^{-2}
\displaystyle =\frac{1}{(x+1)^2} \qquad ...(2)
Also, from (1),
\displaystyle \left(\frac{dy}{dx}\right)^2=\left(-\frac{1}{x+1}\right)^2
\displaystyle =\frac{1}{(x+1)^2} \qquad ...(3)
From (2) and (3),
\displaystyle \frac{d^2y}{dx^2}=\left(\frac{dy}{dx}\right)^2
Hence,
\boxed{\displaystyle \frac{d^2y}{dx^2}=\left(\frac{dy}{dx}\right)^2}
Question 17: Second Order Derivative
17. If y=(\tan^{-1}x)^2, show that (x^2+1)^2y_2+2x(x^2+1)y_1=2.
Solution
Given,
y=(\tan^{-1}x)^2
Differentiating w.r.t. x, we get
\displaystyle y_1=\frac{dy}{dx}=2\tan^{-1}x\cdot\frac{1}{1+x^2}
\displaystyle =\frac{2\tan^{-1}x}{1+x^2} \qquad ...(1)
Multiplying both sides of (1) by (1+x^2), we get
\displaystyle (1+x^2)y_1=2\tan^{-1}x \qquad ...(2)
Differentiating (2) w.r.t. x, we get
\displaystyle \frac{d}{dx}\Big((1+x^2)y_1\Big)=\frac{d}{dx}\Big(2\tan^{-1}x\Big)
Using the Product Rule,
\displaystyle (1+x^2)y_2+2xy_1=\frac{2}{1+x^2}
Multiplying both sides by (1+x^2), we get
\displaystyle (1+x^2)^2y_2+2x(1+x^2)y_1=2
Hence,
\boxed{\displaystyle (x^2+1)^2y_2+2x(x^2+1)y_1=2}
Common Mistakes to Avoid
- Stopping after the first derivative: Many students find \frac{dy}{dx} correctly but forget that the question asks for \frac{d^2y}{dx^2}. Always differentiate once more to obtain the second derivative.
- Differentiating the original function again: The second derivative is obtained by differentiating \frac{dy}{dx}, not by differentiating the original function y a second time independently.
- Making mistakes in the first derivative: Any error in \frac{dy}{dx} will automatically affect \frac{d^2y}{dx^2}. Therefore, simplify and verify the first derivative before proceeding further.
- Forgetting the Chain Rule: Functions such as \sin(\log x), \log(\log x) and (\tan^{-1}x)^2 require the Chain Rule. Always multiply by the derivative of the inner function.
- Making errors while differentiating inverse trigonometric functions: Remember that \frac{d}{dx}(\sin^{-1}x)=\frac{1}{\sqrt{1-x^2}}, \frac{d}{dx}(\cos^{-1}x)=-\frac{1}{\sqrt{1-x^2}} and \frac{d}{dx}(\tan^{-1}x)=\frac{1}{1+x^2}.
- Forgetting the Product Rule: Expressions such as x\cos x, x^3\log x, e^x\sin5x and e^{6x}\cos3x require the Product Rule.
- Making sign errors: Negative signs are frequently missed while differentiating \cos x, \cos(\log x), \cos^{-1}x and exponential functions with negative exponents.
- Using the Quotient Rule incorrectly: When differentiating expressions such as \frac{\cos(\log x)}{x}, apply the Quotient Rule carefully and simplify the result step by step.
- Not expressing the answer in the required form: In questions such as “find \frac{d^2y}{dx^2} in terms of y alone”, substitute back using the given relation and eliminate x from the final answer.
- Ignoring useful substitutions in proof questions: In questions involving y_1 and y_2, it is often easier to manipulate the first derivative and then differentiate again, rather than expanding the second derivative directly.
Continue Learning
After completing Exercise 5.7 of Continuity and Differentiability, you should now be familiar with finding second order derivatives of algebraic, trigonometric, inverse trigonometric, exponential and logarithmic functions. You have also seen how the Product Rule, Quotient Rule, Chain Rule and implicit differentiation can be applied while obtaining higher-order derivatives.
To strengthen your understanding further, make sure that you revise:
- First and second order derivatives
- Derivatives of algebraic, trigonometric and inverse trigonometric functions
- Derivatives of exponential and logarithmic functions
- The Chain Rule for composite functions
- The Product Rule and Quotient Rule
- Implicit differentiation
- Parametric differentiation and the formula \displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}
- Second derivatives in parametric form
- Simplification using trigonometric identities
- Proof-based questions involving y, y_1 and y_2
Explore More
Try solving a variety of second order derivative problems on your own. First find \frac{dy}{dx} carefully and then differentiate it again to obtain \frac{d^2y}{dx^2}. Practise questions involving algebraic, trigonometric, inverse trigonometric, exponential and logarithmic functions. Regular practice will help you apply the Chain Rule, Product Rule and Quotient Rule correctly and improve your accuracy in finding higher-order derivatives.
All the best and keep learning 👍
