Exercise 5.3

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Q.1 Find $\dfrac{dy}{dx}$ if

$$ 2x+3y=\sin x. $$

Solution:

Given,

$$ 2x+3y=\sin x. $$

Differentiating both sides with respect to $x$,

$$ \frac{d}{dx}(2x)+\frac{d}{dx}(3y) = \frac{d}{dx}(\sin x). $$

Using

$$ \frac{d}{dx}(y)=\frac{dy}{dx}, $$

we get

$$ 2+3\frac{dy}{dx}=\cos x. $$

Therefore,

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

Answer:

$$ \boxed{\frac{dy}{dx}=\frac{\cos x-2}{3}.} $$

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Q.2 Find $\dfrac{dy}{dx}$ if

$$ 2x+3y=\sin y. $$

Solution:

Given,

$$ 2x+3y=\sin y. $$

Differentiating both sides with respect to $x$,

$$ \frac{d}{dx}(2x)+\frac{d}{dx}(3y) = \frac{d}{dx}(\sin y). $$

Using the chain rule,

$$ \frac{d}{dx}(\sin y)=\cos y\frac{dy}{dx}, $$

we obtain

$$ 2+3\frac{dy}{dx} = \cos y\frac{dy}{dx}. $$

Collecting the terms containing $\dfrac{dy}{dx}$,

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

Hence,

$$ \boxed{\frac{dy}{dx}=\frac{-2}{\,3-\cos y\,}.} $$

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Q.3 Find $\dfrac{dy}{dx}$ if

$$ ax+by^2=\cos y, $$

where $a$ and $b$ are constants.

Solution:

Given,

$$ ax+by^2=\cos y. $$

Differentiating both sides with respect to $x$,

$$ \frac{d}{dx}(ax)+\frac{d}{dx}(by^2) = \frac{d}{dx}(\cos y). $$

Using the chain rule,

$$ a+2by\frac{dy}{dx} = -\sin y\frac{dy}{dx}. $$

Collecting the terms containing $\dfrac{dy}{dx}$,

$$ \begin{aligned} 2by\frac{dy}{dx} +\sin y\frac{dy}{dx} &=-a,\\[2mm] (2by+\sin y)\frac{dy}{dx} &=-a. \end{aligned} $$

Therefore,

$$ \boxed{\frac{dy}{dx}=\frac{-a}{\,2by+\sin y\,}.} $$

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Q.4 Find $\dfrac{dy}{dx}$ if

$$ xy+y^2=\tan x+y. $$

Solution:

Given,

$$ xy+y^2=\tan x+y. $$

Differentiating both sides with respect to $x$,

$$ \frac{d}{dx}(xy) + \frac{d}{dx}(y^2) = \frac{d}{dx}(\tan x) + \frac{d}{dx}(y). $$

Using the product rule,

$$ \frac{d}{dx}(xy) = x\frac{dy}{dx}+y, $$

we get

$$ x\frac{dy}{dx} +y + 2y\frac{dy}{dx} = \sec^2x + \frac{dy}{dx}. $$

Collect the terms containing $\dfrac{dy}{dx}$,

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

Hence,

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

Answer:

$$ \boxed{\frac{dy}{dx}=\frac{\sec^2x-y}{x+2y-1}.} $$

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Q.5 Find $\dfrac{dy}{dx}$ if

$$ x^2+xy+y^2=100. $$

Solution:

Given,

$$ x^2+xy+y^2=100. $$

Differentiating both sides with respect to $x$,

$$ \frac{d}{dx}(x^2) +\frac{d}{dx}(xy) +\frac{d}{dx}(y^2) = \frac{d}{dx}(100). $$

Using the product rule,

$$ \frac{d}{dx}(xy) = x\frac{dy}{dx}+y, $$

we get

$$ 2x+x\frac{dy}{dx}+y+2y\frac{dy}{dx}=0. $$

Collecting the terms containing $\dfrac{dy}{dx}$,

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

Hence,

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

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Q.6 Find $\dfrac{dy}{dx}$ if

$$ x^3+x^2y+xy^2+y^3=81. $$

Solution:

Given,

$$ x^3+x^2y+xy^2+y^3=81. $$

Differentiating both sides with respect to $x$,

$$ \frac{d}{dx}(x^3) +\frac{d}{dx}(x^2y) +\frac{d}{dx}(xy^2) +\frac{d}{dx}(y^3) = 0. $$

Using the product rule,

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

Substituting,

$$ 3x^2+2xy+x^2\frac{dy}{dx} +y^2 +2xy\frac{dy}{dx} +3y^2\frac{dy}{dx} =0. $$

Collecting the terms containing $\dfrac{dy}{dx}$,

$$ \begin{aligned} (x^2+2xy+3y^2)\frac{dy}{dx} &=-(3x^2+2xy+y^2). \end{aligned} $$

Therefore,

$$ \boxed{ \frac{dy}{dx} = \frac{-(3x^2+2xy+y^2)} {x^2+2xy+3y^2}. } $$

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Q.7 Find $\dfrac{dy}{dx}$ if

$$ \sin^2y+\cos(xy)=k, $$

where $k$ is a constant.

Solution:

Given,

$$ \sin^2y+\cos(xy)=k. $$

Differentiating both sides with respect to $x$,

$$ \frac{d}{dx}(\sin^2y) + \frac{d}{dx}(\cos(xy)) = 0. $$

Using the chain rule,

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

Hence,

$$ 2\sin y\cos y\frac{dy}{dx} - \sin(xy)\left(x\frac{dy}{dx}+y\right) =0. $$

Expanding,

$$ 2\sin y\cos y\frac{dy}{dx} - x\sin(xy)\frac{dy}{dx} - y\sin(xy) =0. $$

Collecting the terms containing $\dfrac{dy}{dx}$,

$$ \begin{aligned} \left(2\sin y\cos y-x\sin(xy)\right)\frac{dy}{dx} &=y\sin(xy). \end{aligned} $$

Therefore,

$$ \boxed{ \frac{dy}{dx} = \frac{y\sin(xy)} {2\sin y\cos y-x\sin(xy)}. } $$

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Q.8 Find $\dfrac{dy}{dx}$ if

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

Solution:

Given,

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

Differentiating both sides with respect to $x$,

$$ \frac{d}{dx}(\sin^2x) + \frac{d}{dx}(\cos^2y) = 0. $$

Using the chain rule,

$$ 2\sin x\cos x - 2\cos y\sin y\frac{dy}{dx} = 0. $$

Therefore,

$$ 2\cos y\sin y\frac{dy}{dx} = 2\sin x\cos x. $$

Dividing both sides by $2\cos y\sin y$,

$$ \begin{aligned} \frac{dy}{dx} &= \frac{\sin x\cos x} {\sin y\cos y}. \end{aligned} $$

Using

$$ 2\sin A\cos A=\sin2A, $$

the derivative may also be written as

$$ \boxed{ \frac{dy}{dx} = \frac{\sin2x}{\sin2y}. } $$

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Q.9 Find $\dfrac{dy}{dx}$ if

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

Solution:

Given,

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

Let

$$ u=\frac{2x}{1+x^2}. $$

Using the chain rule,

$$ \frac{dy}{dx} = \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}. $$

Step 1: Find $\dfrac{du}{dx}$.

Using the quotient rule,

$$ \begin{aligned} \frac{du}{dx} &= \frac{(1+x^2)(2)-2x(2x)} {(1+x^2)^2}\\[2mm] &= \frac{2+2x^2-4x^2} {(1+x^2)^2}\\[2mm] &= \frac{2(1-x^2)} {(1+x^2)^2}. \end{aligned} $$

Step 2: Find $1-u^2$.

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

Therefore,

$$ \sqrt{1-u^2} = \frac{|1-x^2|}{1+x^2}. $$

Hence,

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

For $|x|<1$, this becomes

$$ \boxed{\frac{dy}{dx}=\frac{2}{1+x^2}.} $$

For $|x|>1$, this becomes

$$ \boxed{\frac{dy}{dx}=-\frac{2}{1+x^2}.} $$

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Q.10 Find $\dfrac{dy}{dx}$ if

Solution:

Given,

$$ y=\tan^{-1}\left(\frac{3x-x^3}{1-3x^2}\right). $$

Using the identity

$$ \tan3\theta = \frac{3\tan\theta-\tan^3\theta} {1-3\tan^2\theta}, $$

put

$$ x=\tan\theta. $$

Then,

$$ \frac{3x-x^3}{1-3x^2} = \tan3\theta. $$

Hence,

$$ y=\tan^{-1}(\tan3\theta). $$

Since

$$ -\frac1{\sqrt3} we have

$$ -\frac{\pi}{6}<\theta<\frac{\pi}{6}, $$

which implies

$$ -\frac{\pi}{2}<3\theta<\frac{\pi}{2}. $$

Therefore,

$$ y=3\theta. $$

Since

$$ \theta=\tan^{-1}x, $$

we obtain

$$ y=3\tan^{-1}x. $$

Differentiating,

$$ \begin{aligned} \frac{dy}{dx} &= 3\cdot\frac{1}{1+x^2}. \end{aligned} $$

Answer:

$$ \boxed{\frac{dy}{dx}=\frac{3}{1+x^2}.} $$

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Q.11 Find $\dfrac{dy}{dx}$ if

Solution:

Given,

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

Using the identity

$$ \cos2\theta=\frac{1-\tan^2\theta}{1+\tan^2\theta}, $$

let

$$ x=\tan\theta. $$

Then,

$$ \frac{1-x^2}{1+x^2}=\cos2\theta. $$

Hence,

$$ y=\cos^{-1}(\cos2\theta). $$

Since $0 $$ 0<\theta<\frac{\pi}{4}, $$

which implies

$$ 0<2\theta<\frac{\pi}{2}. $$

Therefore,

$$ y=2\theta=2\tan^{-1}x. $$

Differentiating both sides with respect to $x$,

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

Answer:

$$ \boxed{\frac{dy}{dx}=\frac{2}{1+x^2}.} $$

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Q.12 Find $\dfrac{dy}{dx}$ if

Solution:

Given,

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

Using the identity

$$ \sin\!\left(\frac{\pi}{2}-2\theta\right) = \cos2\theta = \frac{1-\tan^2\theta}{1+\tan^2\theta}, $$

let

$$ x=\tan\theta. $$

Then,

$$ \frac{1-x^2}{1+x^2} = \sin\!\left(\frac{\pi}{2}-2\theta\right). $$

Since $0 $$ 0<\theta<\frac{\pi}{4}, $$

and therefore,

$$ 0<\frac{\pi}{2}-2\theta<\frac{\pi}{2}. $$

Hence,

$$ y=\frac{\pi}{2}-2\theta =\frac{\pi}{2}-2\tan^{-1}x. $$

Differentiating,

$$ \begin{aligned} \frac{dy}{dx} &= 0-2\cdot\frac{1}{1+x^2}\\ &= -\frac{2}{1+x^2}. \end{aligned} $$

Answer:

$$ \boxed{\frac{dy}{dx}=-\frac{2}{1+x^2}.} $$

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Q.13 Find $\dfrac{dy}{dx}$ if

Solution:

Given,

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

Using the identity

$$ \sin2\theta = \frac{2\tan\theta}{1+\tan^2\theta}, $$

let

$$ x=\tan\theta. $$

Then,

$$ \frac{2x}{1+x^2} = \sin2\theta. $$

Also,

$$ -1 which gives

$$ -\frac{\pi}{2}<2\theta<\frac{\pi}{2}. $$

Since the principal value of $\cos^{-1}$ lies in $[0,\pi]$,

$$ \cos^{-1}(\sin2\theta) = \frac{\pi}{2}-2\theta. $$

Therefore,

$$ y = \frac{\pi}{2} - 2\tan^{-1}x. $$

Differentiating,

$$ \begin{aligned} \frac{dy}{dx} &= 0-2\cdot\frac{1}{1+x^2}\\ &= -\frac{2}{1+x^2}. \end{aligned} $$

Answer:

$$ \boxed{\frac{dy}{dx}=-\frac{2}{1+x^2}.} $$

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Q.14 Find $\dfrac{dy}{dx}$ if

Solution:

Given,

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

Using the identity

$$ \sin2\theta=2\sin\theta\cos\theta, $$

let

$$ x=\sin\theta. $$

Then,

$$ \sqrt{1-x^2}=\cos\theta, $$

since

$$ -\frac{1}{\sqrt2} and therefore $\cos\theta>0$.

Hence,

$$ 2x\sqrt{1-x^2} = 2\sin\theta\cos\theta = \sin2\theta. $$

Thus,

$$ y=\sin^{-1}(\sin2\theta). $$

Since

$$ -\frac{\pi}{2}<2\theta<\frac{\pi}{2}, $$

we obtain

$$ y=2\theta=2\sin^{-1}x. $$

Differentiating both sides,

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

Answer:

$$ \boxed{\frac{dy}{dx} = \frac{2}{\sqrt{1-x^2}}.} $$

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Q.15 Find $\dfrac{dy}{dx}$ if

Solution:

Given,

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

Using the identity

$$ \sec2\theta=\frac{1}{1-2\sin^2\theta}, $$

let

$$ x=\sin\theta. $$

Then,

$$ 2x^2-1 = 2\sin^2\theta-1 = -(1-2\sin^2\theta) = -\cos2\theta. $$

Therefore,

$$ \frac{1}{2x^2-1} = -\sec2\theta = \sec(\pi-2\theta). $$

Hence,

$$ y=\sec^{-1}\big(\sec(\pi-2\theta)\big). $$

Since

$$ 0 we have

$$ 0<\theta<\frac{\pi}{4}, $$

which gives

$$ \frac{\pi}{2}<\pi-2\theta<\pi. $$

This lies in the principal range of the inverse secant function. Therefore,

$$ y=\pi-2\theta = \pi-2\sin^{-1}x. $$

Differentiating both sides,

$$ \begin{aligned} \frac{dy}{dx} &= 0 - 2\cdot\frac{1}{\sqrt{1-x^2}}\\[2mm] &= -\frac{2}{\sqrt{1-x^2}}. \end{aligned} $$

Answer:

$$ \boxed{\frac{dy}{dx} = -\frac{2}{\sqrt{1-x^2}}.} $$

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