Exercise 5.7

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Q.1 Find the second order derivative of

$$ y=x^2+3x+2. $$

Solution:

Given,

$$ y=x^2+3x+2. $$

Step 1: Find the first derivative.

Differentiating with respect to $x$,

$$ \begin{aligned} \frac{dy}{dx} &=\frac{d}{dx}(x^2)+\frac{d}{dx}(3x)+\frac{d}{dx}(2)\\[2mm] &=2x+3. \end{aligned} $$

Step 2: Find the second derivative.

Differentiate the first derivative once again.

$$ \begin{aligned} \frac{d^2y}{dx^2} &=\frac{d}{dx}(2x+3)\\[2mm] &=2. \end{aligned} $$

Answer:

$$ \boxed{\frac{d^2y}{dx^2}=2.} $$

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Q.2 Find the second order derivative of

$$ y=x^{20}. $$

Solution:

Given,

$$ y=x^{20}. $$

Step 1: Find the first derivative.

$$ \begin{aligned} \frac{dy}{dx} &=20x^{19}. \end{aligned} $$

Step 2: Find the second derivative.

$$ \begin{aligned} \frac{d^2y}{dx^2} &=\frac{d}{dx}(20x^{19})\\[2mm] &=20\times19x^{18}\\[2mm] &=380x^{18}. \end{aligned} $$

Answer:

$$ \boxed{\frac{d^2y}{dx^2}=380x^{18}.} $$

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Q.3 Find the second order derivative of

$$ y=x\cos x. $$

Solution:

Given,

$$ y=x\cos x. $$

Step 1: Find the first derivative.

Using the product rule,

$$ \begin{aligned} \frac{dy}{dx} &=x\frac{d}{dx}(\cos x)+\cos x\frac{d}{dx}(x)\\[2mm] &=-x\sin x+\cos x. \end{aligned} $$

Step 2: Find the second derivative.

Differentiate the first derivative again.

$$ \begin{aligned} \frac{d^2y}{dx^2} &=\frac{d}{dx}(-x\sin x)+\frac{d}{dx}(\cos x). \end{aligned} $$

Using the product rule,

$$ \begin{aligned} \frac{d}{dx}(-x\sin x) &=-\left(\sin x+x\cos x\right). \end{aligned} $$

Also,

$$ \frac{d}{dx}(\cos x)=-\sin x. $$

Therefore,

$$ \begin{aligned} \frac{d^2y}{dx^2} &=-\sin x-x\cos x-\sin x\\[2mm] &=-2\sin x-x\cos x. \end{aligned} $$

Answer:

$$ \boxed{\frac{d^2y}{dx^2}=-x\cos x-2\sin x.} $$

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Q.4 Find the second order derivative of

$$ y=\log x. $$

Solution:

Given,

$$ y=\log x. $$

Step 1: Find the first derivative.

$$ \begin{aligned} \frac{dy}{dx} &=\frac{d}{dx}(\log x)\\[2mm] &=\frac1x. \end{aligned} $$

Step 2: Find the second derivative.

$$ \begin{aligned} \frac{d^2y}{dx^2} &=\frac{d}{dx}\left(\frac1x\right)\\[2mm] &=\frac{d}{dx}(x^{-1})\\[2mm] &=-x^{-2}\\[2mm] &=-\frac1{x^2}. \end{aligned} $$

Answer:

$$ \boxed{\frac{d^2y}{dx^2}=-\frac1{x^2}.} $$

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Q.5 Find the second order derivative of

$$ y=x^3\log x. $$

Solution:

Given,

$$ y=x^3\log x. $$

Step 1: Find the first derivative.

Using the product rule,

$$ \begin{aligned} \frac{dy}{dx} &=x^3\cdot\frac1x+\log x\cdot3x^2\\[2mm] &=x^2+3x^2\log x\\[2mm] &=x^2(1+3\log x). \end{aligned} $$

Step 2: Find the second derivative.

Again using the product rule,

$$ \begin{aligned} \frac{d^2y}{dx^2} &=2x(1+3\log x) +x^2\left(\frac3x\right)\\[2mm] &=2x+6x\log x+3x\\[2mm] &=5x+6x\log x. \end{aligned} $$

Answer:

$$ \boxed{\frac{d^2y}{dx^2}=5x+6x\log x=x(5+6\log x).} $$

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Q.6 Find the second order derivative of

$$ y=e^x\sin5x. $$

Solution:

Given,

$$ y=e^x\sin5x. $$

Step 1: Find the first derivative.

Using the product rule,

$$ \begin{aligned} \frac{dy}{dx} &=e^x\sin5x+e^x(5\cos5x)\\[2mm] &=e^x(\sin5x+5\cos5x). \end{aligned} $$

Step 2: Find the second derivative.

Again using the product rule,

$$ \begin{aligned} \frac{d^2y}{dx^2} &=e^x(\sin5x+5\cos5x)\\ &\quad+e^x(5\cos5x-25\sin5x)\\[2mm] &=e^x(10\cos5x-24\sin5x). \end{aligned} $$

Answer:

$$ \boxed{\frac{d^2y}{dx^2}=e^x\left(10\cos5x-24\sin5x\right).} $$

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Q.7 Find the second order derivative of

$$ y=e^{6x}\cos3x. $$

Solution:

Given,

$$ y=e^{6x}\cos3x. $$

Step 1: Find the first derivative.

Using the product rule,

$$ \begin{aligned} \frac{dy}{dx} &= e^{6x}\frac{d}{dx}(\cos3x) + \cos3x\frac{d}{dx}(e^{6x})\\[2mm] &= e^{6x}(-3\sin3x) + 6e^{6x}\cos3x\\[2mm] &= e^{6x}(6\cos3x-3\sin3x). \end{aligned} $$

Step 2: Find the second derivative.

Again using the product rule,

$$ \begin{aligned} \frac{d^2y}{dx^2} &= 6e^{6x}(6\cos3x-3\sin3x)\\ &\quad + e^{6x}(-18\sin3x-9\cos3x)\\[2mm] &= e^{6x} (27\cos3x-36\sin3x). \end{aligned} $$

Answer:

$$ \boxed{ \frac{d^2y}{dx^2} = e^{6x}(27\cos3x-36\sin3x). } $$

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Q.8 Find the second order derivative of

$$ y=\tan^{-1}x. $$

Solution:

Given,

$$ y=\tan^{-1}x. $$

Step 1: Find the first derivative.

$$ \begin{aligned} \frac{dy}{dx} &= \frac{1}{1+x^2}. \end{aligned} $$

Step 2: Find the second derivative.

$$ \begin{aligned} \frac{d^2y}{dx^2} &= \frac{d}{dx}(1+x^2)^{-1}\\[2mm] &= -(1+x^2)^{-2}(2x)\\[2mm] &= -\frac{2x}{(1+x^2)^2}. \end{aligned} $$

Answer:

$$ \boxed{ \frac{d^2y}{dx^2} = -\frac{2x}{(1+x^2)^2}. } $$

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Q.9 Find the second order derivative of

$$ y=\log(\log x). $$

Solution:

Given,

$$ y=\log(\log x). $$

Step 1: Find the first derivative.

$$ \begin{aligned} \frac{dy}{dx} &= \frac{1}{\log x}\cdot\frac1x\\[2mm] &= \frac{1}{x\log x}. \end{aligned} $$

Step 2: Find the second derivative.

Write

$$ \frac{dy}{dx} = (x\log x)^{-1}. $$

Differentiating,

$$ \begin{aligned} \frac{d^2y}{dx^2} &= -(x\log x)^{-2} \cdot \frac{d}{dx}(x\log x). \end{aligned} $$

Now,

$$ \begin{aligned} \frac{d}{dx}(x\log x) &= \log x+1. \end{aligned} $$

Hence,

$$ \begin{aligned} \frac{d^2y}{dx^2} &= -\frac{\log x+1} {x^2(\log x)^2}. \end{aligned} $$

Answer:

$$ \boxed{ \frac{d^2y}{dx^2} = -\frac{\log x+1} {x^2(\log x)^2}. } $$

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Q.10 Find the second order derivative of

$$ y=\sin(\log x). $$

Solution:

Given,

$$ y=\sin(\log x). $$

Step 1: Find the first derivative.

Using the chain rule,

$$ \begin{aligned} \frac{dy}{dx} &= \cos(\log x)\cdot\frac{d}{dx}(\log x)\\[2mm] &= \frac{\cos(\log x)}{x}. \end{aligned} $$

Step 2: Find the second derivative.

Using the product rule,

$$ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(x^{-1}\cos(\log x)\right). $$

Therefore,

$$ \begin{aligned} \frac{d^2y}{dx^2} &= (-x^{-2})\cos(\log x) + x^{-1} \left[ -\sin(\log x)\cdot\frac1x \right]\\[2mm] &= -\frac{\cos(\log x)}{x^2} - \frac{\sin(\log x)}{x^2}. \end{aligned} $$

Hence,

$$ \boxed{ \frac{d^2y}{dx^2} = -\frac{\sin(\log x)+\cos(\log x)}{x^2}. } $$

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Q.11 If

$$ y=5\cos x-3\sin x, $$

prove that

$$ \frac{d^2y}{dx^2}+y=0. $$

Solution:

Given,

$$ y=5\cos x-3\sin x. $$

Step 1: Find the first derivative.

$$ \begin{aligned} \frac{dy}{dx} &= 5(-\sin x)-3(\cos x)\\[2mm] &= -5\sin x-3\cos x. \end{aligned} $$

Step 2: Find the second derivative.

$$ \begin{aligned} \frac{d^2y}{dx^2} &= -5\cos x+3\sin x. \end{aligned} $$

Step 3: Compute $\dfrac{d^2y}{dx^2}+y$.

$$ \begin{aligned} \frac{d^2y}{dx^2}+y &= (-5\cos x+3\sin x) + (5\cos x-3\sin x)\\[2mm] &= 0. \end{aligned} $$

Hence,

$$ \boxed{ \frac{d^2y}{dx^2}+y=0. } $$

Thus, the required result is proved.

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Q.12 If

$$ y=\cos^{-1}x, $$

find

$$ \frac{d^2y}{dx^2} $$

in terms of $y$ alone.

Solution:

Given,

$$ y=\cos^{-1}x. $$

Step 1: Find the first derivative.

$$ \begin{aligned} \frac{dy}{dx} &= -\frac{1}{\sqrt{1-x^2}}. \end{aligned} $$

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Step 2: Find the second derivative.

$$ \begin{aligned} \frac{d^2y}{dx^2} &= -\frac{d}{dx}(1-x^2)^{-\frac12}\\[2mm] &= -\left(-\frac12\right) (1-x^2)^{-\frac32} (-2x)\\[2mm] &= -\frac{x}{(1-x^2)^{3/2}}. \end{aligned} $$

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Step 3: Express the result in terms of $y$.

Since

$$ y=\cos^{-1}x, $$

we have

$$ x=\cos y. $$

Also,

$$ 1-x^2 = 1-\cos^2y = \sin^2y. $$

Therefore,

$$ (1-x^2)^{3/2} = (\sin^2y)^{3/2} = \sin^3y, $$

because $0\le y\le\pi$ and hence $\sin y\ge0$.

Substituting,

$$ \boxed{ \frac{d^2y}{dx^2} = -\frac{\cos y}{\sin^3y} = -\cot y\,\csc^2y. } $$

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Q.13 If

$$ y=3\cos(\log x)+4\sin(\log x), $$

show that

$$ x^2y_2+xy_1+y=0, $$

where

$$ y_1=\frac{dy}{dx}, \qquad y_2=\frac{d^2y}{dx^2}. $$

Solution:

Given,

$$ y=3\cos(\log x)+4\sin(\log x). $$

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Step 1: Find the first derivative.

$$ \begin{aligned} y_1 &= 3\left(-\sin(\log x)\right)\frac1x + 4\cos(\log x)\frac1x\\[2mm] &= \frac{-3\sin(\log x)+4\cos(\log x)}{x}. \end{aligned} $$

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Step 2: Find the second derivative.

Using the product rule,

$$ y_1= x^{-1} \left( -3\sin(\log x)+4\cos(\log x) \right). $$

Differentiating,

$$ \begin{aligned} y_2 &= -\frac1{x^2} \left( -3\sin(\log x)+4\cos(\log x) \right) \\[2mm] &\quad + \frac1x \left( -\frac{3\cos(\log x)}x - \frac{4\sin(\log x)}x \right) \\[2mm] &= \frac{ 3\sin(\log x) - 4\cos(\log x) - 3\cos(\log x) - 4\sin(\log x) }{x^2} \\[2mm] &= -\frac{ \sin(\log x)+7\cos(\log x) }{x^2}. \end{aligned} $$

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Step 3: Evaluate $x^2y_2+xy_1+y$.

Now,

$$ x^2y_2 = -\sin(\log x)-7\cos(\log x), $$ $$ xy_1 = -3\sin(\log x)+4\cos(\log x), $$

and

$$ y = 3\cos(\log x)+4\sin(\log x). $$

Therefore,

$$ \begin{aligned} x^2y_2+xy_1+y &= \left(-\sin-7\cos\right) +\left(-3\sin+4\cos\right) +\left(3\cos+4\sin\right) \\[2mm] &= 0. \end{aligned} $$

Hence,

$$ \boxed{ x^2y_2+xy_1+y=0. } $$

Thus, the required result is proved.

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Q.14 If

$$ y=Ae^{mx}+Be^{nx}, $$

show that

$$ \frac{d^2y}{dx^2}-(m+n)\frac{dy}{dx}+mny=0. $$

Solution:

Given,

$$ y=Ae^{mx}+Be^{nx}, $$

where $A$, $B$, $m$ and $n$ are constants.

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Step 1: Find the first derivative.

Differentiating with respect to $x$,

$$ \begin{aligned} \frac{dy}{dx} &= Am e^{mx} + Bn e^{nx}. \end{aligned} $$

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Step 2: Find the second derivative.

Differentiating once again,

$$ \begin{aligned} \frac{d^2y}{dx^2} &= Am^2e^{mx} + Bn^2e^{nx}. \end{aligned} $$

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Step 3: Evaluate

$$ \frac{d^2y}{dx^2} -(m+n)\frac{dy}{dx} +mny. $$

Substituting the values of $y$, $\dfrac{dy}{dx}$ and $\dfrac{d^2y}{dx^2}$,

$$ \begin{aligned} &\frac{d^2y}{dx^2} -(m+n)\frac{dy}{dx} +mny \\[2mm] ={}& \left( Am^2e^{mx} + Bn^2e^{nx} \right) \\[2mm] &-(m+n) \left( Ame^{mx} +Bne^{nx} \right) \\[2mm] &+mn \left( Ae^{mx} + Be^{nx} \right). \end{aligned} $$

Grouping the terms containing $Ae^{mx}$,

$$ Ae^{mx} \left[ m^2-m(m+n)+mn \right]. $$

Since

$$ m^2-m(m+n)+mn = m^2-m^2-mn+mn = 0, $$

the coefficient of $Ae^{mx}$ is zero.

Similarly, grouping the terms containing $Be^{nx}$,

$$ Be^{nx} \left[ n^2-n(m+n)+mn \right]. $$

Now,

$$ n^2-n(m+n)+mn = n^2-mn-n^2+mn = 0. $$

Hence,

$$ \boxed{ \frac{d^2y}{dx^2} -(m+n)\frac{dy}{dx} +mny =0. } $$

Thus, the required result is proved.

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Q.15 If

$$ y=500e^{7x}+600e^{-7x}, $$

show that

$$ \frac{d^2y}{dx^2}=49y. $$

Solution:

Given,

$$ y=500e^{7x}+600e^{-7x}. $$

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Step 1: Find the first derivative.

$$ \begin{aligned} \frac{dy}{dx} &= 500(7e^{7x}) + 600(-7e^{-7x}) \\[2mm] &= 3500e^{7x} - 4200e^{-7x}. \end{aligned} $$

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Step 2: Find the second derivative.

$$ \begin{aligned} \frac{d^2y}{dx^2} &= 3500(7e^{7x}) - 4200(-7e^{-7x}) \\[2mm] &= 24500e^{7x} + 29400e^{-7x}. \end{aligned} $$

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Step 3: Evaluate $49y$.

$$ \begin{aligned} 49y &= 49 \left( 500e^{7x} + 600e^{-7x} \right) \\[2mm] &= 24500e^{7x} + 29400e^{-7x}. \end{aligned} $$

Therefore,

$$ \boxed{ \frac{d^2y}{dx^2} = 49y. } $$

Hence proved.

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Q.16 If

$$ e^y(x+1)=1, $$

show that

$$ \frac{d^2y}{dx^2}=\left(\frac{dy}{dx}\right)^2. $$

Solution:

Given,

$$ e^y(x+1)=1. $$

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Step 1: Find the first derivative.

Differentiate both sides with respect to $x$ using the product rule.

$$ \begin{aligned} \frac{d}{dx} \left[e^y(x+1)\right] &= 0. \end{aligned} $$

Applying the product rule,

$$ e^y\frac{dy}{dx}(x+1) + e^y = 0. $$

Factorising $e^y$,

$$ e^y \left[ (x+1)\frac{dy}{dx}+1 \right] =0. $$

Since $e^y\neq0$,

$$ (x+1)\frac{dy}{dx}+1=0. $$

Hence,

$$ \boxed{ \frac{dy}{dx} = -\frac1{x+1}. } $$

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Step 2: Find the second derivative.

Differentiating once again,

$$ \begin{aligned} \frac{d^2y}{dx^2} &= \frac{d}{dx} \left( -\frac1{x+1} \right) \\[2mm] &= \frac1{(x+1)^2}. \end{aligned} $$

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Step 3: Evaluate

$$ \left(\frac{dy}{dx}\right)^2. $$

From the first derivative,

$$ \begin{aligned} \left(\frac{dy}{dx}\right)^2 &= \left( -\frac1{x+1} \right)^2 \\[2mm] &= \frac1{(x+1)^2}. \end{aligned} $$

Therefore,

$$ \boxed{ \frac{d^2y}{dx^2} = \left( \frac{dy}{dx} \right)^2. } $$

Hence proved.

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Q.17 If

$$ y=\left(\tan^{-1}x\right)^2, $$

show that

$$ (x^2+1)^2y_2+2x(x^2+1)y_1=2, $$

where

$$ y_1=\frac{dy}{dx}, \qquad y_2=\frac{d^2y}{dx^2}. $$

Solution:

Given,

$$ y=\left(\tan^{-1}x\right)^2. $$

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Step 1: Find the first derivative.

Using the chain rule,

$$ \begin{aligned} y_1 &= 2\tan^{-1}x \cdot \frac1{1+x^2} \\[2mm] &= \frac{2\tan^{-1}x}{1+x^2}. \end{aligned} $$

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Step 2: Find the second derivative.

Using the quotient rule,

$$ \begin{aligned} y_2 &= \frac{ (1+x^2)\left(\dfrac{2}{1+x^2}\right) - 2\tan^{-1}x(2x) } {(1+x^2)^2} \\[2mm] &= \frac{ 2-4x\tan^{-1}x/(1+x^2) } {1+x^2}. \end{aligned} $$

Writing in a simplified form,

$$ \boxed{ y_2 = \frac{2}{(1+x^2)^2} - \frac{4x\tan^{-1}x}{(1+x^2)^3}. } $$

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Step 3: Evaluate

$$ (1+x^2)^2y_2+2x(1+x^2)y_1. $$

Substituting the values of $y_1$ and $y_2$,

$$ \begin{aligned} &(1+x^2)^2 \left[ \frac{2}{(1+x^2)^2} - \frac{4x\tan^{-1}x}{(1+x^2)^3} \right] \\[2mm] &\quad+ 2x(1+x^2) \left[ \frac{2\tan^{-1}x}{1+x^2} \right] \\[2mm] &= 2 - \frac{4x\tan^{-1}x}{1+x^2} + 4x\tan^{-1}x \cdot \frac1{1+x^2} \\[2mm] &= 2. \end{aligned} $$

Hence,

$$ \boxed{ (x^2+1)^2y_2+2x(x^2+1)y_1=2. } $$

Thus, the required result is proved.

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