Calculus Solutions Ex#3.1

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Q.01 (a) Find $\frac{d y}{d x}$ by differentiating implicitly.
Solution:
$$ y=x+x y-2 x^{3}=2—(1) $$ Differentiating w.r.t $x$ $$ \begin{aligned} & \frac{d}{d x}\left[x+x y-2 x^{3}\right]=\frac{d}{d x}[2] \\ \\ & \frac{d}{d x}[x]+\frac{d}{d x}[x y]-2 \frac{d}{d x}\left[x^{3}\right]=0 \\ \\ & 1+x \frac{d y}{d x}+y \frac{d}{d x}[x]-2\left(3 x^{3-1}\right)=0 \\ \\ & 1+\frac{x d y}{d x}+y-6 x^{2}=0 \\ \\ & x \frac{d y}{d x}=6 x^{2}-y-1 \\ \\ & \frac{d y}{d x}=\frac{6 x^{2}-y-1}{x} \end{aligned} $$ (b) Solving (1) for $y$ $$ \begin{aligned} & xy=2+2 x^{3}-x \\ \\ & y=\frac{2}{x}+\frac{2 x^{3}}{x}-\frac{x}{x} \\ \\ & y=\frac{2}{x}+2 x^{2}-1 \end{aligned} $$ Differentiating w.r.t $x$ $$ \begin{aligned} & \frac{d y}{d x}=\frac{d}{d x}\left[\frac{2}{x}+2 x^{2}-1\right]$$ $$ \frac{d y}{d x}=\frac{d}{d x}\left[2 x^{-1}+2 x^{2}-1\right] $$ $$ \frac{d y}{d x}=[2(-1 x^{-1-1})+2(2 x^{2-1})-\frac{d}{d x}[1]\right. $$ $$ \frac{d y}{d x}=-2 x^{-2}+4 x-0 $$ $$ \frac{d y}{d x}=-\frac{2}{x^{2}}+4 x \end{aligned} $$ (c) From part (a) $$ \begin{aligned} & \frac{d y}{d x}=\frac{6 x^{2}-y-1}{x} \\ \\ & \frac{d y}{d x}=\frac{6 x^{2}}{x}-\frac{y}{x}-\frac{1}{x} \\ \\ & \frac{d y}{d x}=6 x-\frac{1}{x} y-\frac{1}{x} \end{aligned} $$ Substituting $y=2 / x+2 x^{2}-1$ (from part(b)) into above equation, we have $$ \begin{aligned} \frac{d y}{d x} & =6 x-\frac{1}{x}\left[\frac{2}{x}+2 x^{2}-1\right]-\frac{1}{x} \\ \\ & =6 x-\frac{2}{x^{2}}-2 x+\frac{1}{x}-\frac{1}{x} \\ \\ \frac{d y}{d x} & =4 x-\frac{2}{x^{2}} \end{aligned} $$
Q.02 Find $\frac{d y}{d x}$ for $ \sqrt{y}-\sin x=2$ by differentiating implicitly.
Solution:
(a) $$ \begin{aligned} & \sqrt{y}-\sin x=2 \\ \\ & y^{1 / 2}-\sin x=2 \end{aligned} $$ Differentiating w.r.t $x$ $$ \begin{aligned} \frac{d}{d x}\left[y^{1 / 2}\right]-\frac{d}{d x}[\sin x]=\frac{d}{d x}[2] \\ \\ \frac{1}{2} y^{\frac{1}{2}-1} \frac{d y}{d x}-\cos x=0 \\ \\ \frac{1}{2} y^{-1/2} \frac{d y}{d x}-\cos x=0 \\ \\ \frac{1}{2 \sqrt{y}} \frac{d y}{d x}=\cos x \\ \\ \frac{d y}{d x}=2 \sqrt{y} \cos x —(1) \end{aligned} $$ (b) solving $ \sqrt{y}=\sin x+2 $ for $y.$ $$ \sqrt{y}=\sin x+2 $$ squaring both sides $$ y=(\sin x+2)^{2} $$ Differentiating w.r.t $x$ $$ \begin{aligned} \frac{d y}{d x}=\frac{d}{d x}[\sin x+2]^{2} \\ \\ \frac{d y}{d x}=2[\sin x+2]^{2-1} \frac{d}{d x}[\sin x+2] \\ \\ \frac{d y}{d x}=2[\sin x+2][\cos x+0] \\ \\ \frac{d y}{d x}=2 \sin x \cos x+4 \cos x \end{aligned} $$ (c) Substituting $\sqrt{y}=\sin x+2$ into (1) (in part (a))

$$ \begin{aligned} \frac{d y}{d x}=2(\sin x+2) \cos x \\ \\ \frac{d y}{d x}=2 \sin x \cos x+4 \cos x . \end{aligned} $$
Q.03 Find $\frac{d y}{d x}$ by implicit differentiation.
Solution:
$$\quad x^{2}+y^{2}=100$$ Differentiating implicitly w.r.t $x$ $$ \begin{aligned} & \frac{d}{d x}\left[x^{2}+y^{2}\right]=\frac{d}{d x}[100] \\ \\ \frac{d}{d x}\left[x^{2}\right]+\frac{d}{d x}\left[y^{2}\right]=0 \end{aligned} $$ $$ \begin{aligned} & 2 x^{2-1}+2 y^{2-1} \frac{d y}{d x}=0 & \quad \begin{array}{l} \text { chain rule because } \\ $y$ \text { is a function of } $x$ \end{array} \\ & 2 x+2 y \frac{d y}{d x}=0 \\ & 2 y \frac{d y}{d x}=-2 x \Rightarrow \frac{d y}{d x}=\frac{-2 x}{2 y}=-x / y \end{aligned} $$