Differentiation (Part-38) | S N De

 

$~5(iv)~~(\cos x)^y=(\sin y)^x$

Sol. $~~(\cos x)^y=(\sin y)^x \\ \text{or,}~~ \log(\cos x)^y=\log(\sin y)^x \\ \text{or,}~~ y \log(\cos x)=x \log(\sin y) \\ \therefore~~\text{Differentiating w.r.t.}~~x,\\~~~\frac{dy}{dx}~\log(\cos x)+y~\frac{d}{dx}(\log \cos x)\\~~~~~~=\frac{d}{dx}(x)~\log(\sin y)+x~\frac{d}{dx}(\log \sin y) \\ \text{or,}~~ \frac{dy}{dx}~\log(\cos x)+y ~\frac{1}{\cos x} \cdot(-\sin x)\\~~~~~~=1 \cdot \log(\sin y)+\frac{1}{\sin y}\cdot \cos y~\frac{dy}{dx} \\ \text{or,}~~  \frac{dy}{dx}~\log(\cos x)-y~\tan x\\~~~~~~=\log(\sin y)+x \cdot\cot y~\frac{dy}{dx} \\ \text{or,}~~ \frac{dy}{dx}~[\log(\cos x)-x\cot y~]\\~~~~~~=\log(\sin y)+x\cot y \\ \text{or,}~~ \frac{dy}{dx}=\frac{\log(\sin y)+x\cot y}{\log(\cos x)-x\cot y}~~\text{(ans.)}$

$~(v)~~e^{xy}-4xy=4$

Sol. $~~e^{xy}-4xy=4 \rightarrow(1) \\ \therefore \frac{d}{dx}(e^{xy}-4xy)=\frac{d}{dx}(4) \\ \text{or,}~~ \frac{d}{dx}(e^{xy})-4\frac{d}{dx}(xy)=0 \\ \text{or,}~~ e^{xy}~\frac{d}{dx}(xy)-4 \cdot \frac{d}{dx}(xy)=0 \\ \text{or,}~~ \frac{d}{dx}(xy)~(e^{xy}-4)=0 \\ \text{or,}~~ \frac{d}{dx}(xy) \cdot 4xy=0~~[\text{By (1)}] \\ \text{or,}~~ \frac{d}{dx}(xy)=0 \rightarrow(2)$

Here,$~~4xy \neq 0~~$ as  $~~4xy=0 \Rightarrow x=0~~\text{or,}~~y=0~$ which does not satisfy the equation $~(1).$

Now, by $~(2)~$, we get,

$~~\frac{d}{dx}(xy)=0 \\ \text{or,}~~x~\frac{dy}{dx}+y \cdot 1=0 \\ \text{or,}~~x~\frac{dy}{dx}=-y \\ \text{or,}~~ \frac{dy}{dx}=-\frac yx~~\text{(ans.)}$

$~(vi)~~x^3y^4=(x+y)^7$

Sol. $~~x^3y^4=(x+y)^7 \\ \therefore \log_e(x^3y^4)=\log_e(x+y)^7 \\ \Rightarrow \log_e x^3+\log_e y^4=7\log_e(x+y) \\ \Rightarrow 3 \log_e x+4\log_ey=7\log_e(x+y) \\ \therefore 3 \frac{d}{dx}(\log_ex)+4\frac{d}{dx}(\log_ey)\\~~~~~~~~~~~~~~~~=7 \frac{d}{dx} [\log(x+y)] \\ \Rightarrow \frac 3x+\frac{4}{y}~\frac{dy}{dx}=7 \cdot \frac{1}{x+y}\left[1+\frac{dy}{dx}\right]\\ \text{or,}~~ \left(\frac 4y-\frac{7}{x+y}\right)\frac{dy}{dx}=\frac{7}{x+y}-\frac 3x\\ \text{or,}~~ \left(\frac{4x+4y-7y}{y(x+y)}\right)~\frac{dy}{dx}=\frac{7x-3x-3y}{x(x+y)}\\ \text{or,}~~  \left(\frac{4x-3y}{y(x+y)}\right)~\frac{dy}{dx}=\frac{4x-3y}{x(x+y)} \\ \text{or,}~~  \frac{4x-3y}{y(x+y)}~\frac{dy}{dx}=\frac{4x-3y}{x(x+y)} \\ \text{or,}~~  \frac{dy}{dx}=\frac{4x-3y}{x(x+y)} \cdot \frac{y(x+y)}{4x-3y} \\ \text{or,}~~  \frac{dy}{dx}=\frac yx~~\text{(ans.)}$ 

$~(vii)~~xy=\tan(x+y)$

Sol. $~~xy=\tan(x+y) \rightarrow(1) \\ ~~~\text{Diff. w.r.t.}~~x, \\ \text{or,}~~ \frac{d}{dx}(xy)=\frac{d}{dx}[\tan(x+y)]  \\ \text{or,}~~ 1 \cdot y+x \cdot \frac{dy}{dx}=\sec^2(x+y)~\frac{d}{dx}(x+y) \\ \text{or,}~~  y+x~\frac{dy}{dx}=\sec^2(x+y)~(1+\frac{dy}{dx})\\ \text{or,}~~  y+x~\frac{dy}{dx}\\~~~=(1+\tan^2(x+y))(1+\frac{dy}{dx})~~[\text{By (1)}] \\ \text{or,}~~  y+x~\frac{dy}{dx}=(1+x^2y^2)(1+\frac{dy}{dx}) \\ \text{or,}~~  y+x~\frac{dy}{dx}=(1+x^2y^2)+\frac{dy}{dx}(1+x^2y^2) \\ \text{or,}~~  y-1-x^2y^2=\frac{dy}{dx}(1+x^2y^2-x) \\ \text{or,}~~  \frac{dy}{dx}=\frac{y-x^2y^2-1}{1+x^2y^2-x}~~\text{(ans.)}$

$~(viii)~~xy+1=\cos(xy)\\~~\text{when}~~x=\pi/2,~y=0.$

Sol. $~~xy+1=\cos(xy) \\ \therefore ~~\frac{d}{dx}(xy+1)=\frac{d}{dx}[\cos(xy)] \\ \text{or,}~~ \frac{d}{dx}(xy)+0=-\sin(xy)~\frac{d}{dx}(xy) \\ \text{or,}~~  1 \cdot y+x~\frac{dy}{dx}=-\sin(xy)\left(1 \cdot y+x~\frac{dy}{dx}\right) \\ \text{or,}~~  x~\frac{dy}{dx}[1+\sin(xy)]=-y~[\sin(xy)+1] \\ \text{or,}~~  \frac{dy}{dx}=-\frac yx \\ \therefore \left[\frac{dy}{dx}\right]_{x=\pi/2 ,~y=0}=-\frac{0}{\pi/2}=0~~\text{(ans.)}$

$~(ix)~~y^y=\sin x$

Sol. $~~y^y=\sin x \\ \therefore \log_e (y^y)=\log_e (\sin x) \\ \text{or,}~~  y \log_e y=\log_e(\sin x) \\ \therefore \frac{d}{dx}(y \log_e y)=\frac{d}{dx}[\log_e(\sin x)]\\ \text{or,}~~  y \cdot \frac 1y~\frac{dy}{dx}+\frac{dy}{dx}~(\log_ey)=\frac{1}{\sin x}\frac{d}{dx}(\sin x)\\ \text{or,}~~  \frac{dy}{dx}+\frac{dy}{dx}~(\log_e y)=\frac{1}{\sin x} \cdot \cos x\\ \text{or,}~~  \frac{dy}{dx}(1+ \log_ey)=\cot x \\ \text{or,}~~  \frac{dy}{dx}=\frac{\cot x}{1+ \log_ey}~~\text{(ans.)}$

$~(x)~~\log(xy)=e^{x+y}+2$

Sol. $~~\log(xy)=e^{x+y}+2 \\ \therefore \frac{d}{dx}[\log(xy)]=\frac{d}{dx}(e^{x+y}+2) \\ \text{or,}~~\frac{1}{xy}~\frac{d}{dx}(xy)=\frac{d}{dx}(e^{x+y})+0 \\ \text{or,}~~ \frac{1}{xy}~\left(1 \cdot y+x~\frac{dy}{dx}\right)\\~~~~~~=e^{x+y}~\left(1+\frac{dy}{dx}\right) \\ \text{or,}~~ \frac{1}{xy} \cdot y+\frac{1}{xy} \cdot x~\frac{dy}{dx}\\~~~=e^{x+y}+e^{x+y}~\frac{dy}{dx} \\ \text{or,}~~ \frac 1x+\frac 1y~\frac{dy}{dx}=e^{x+y}+e^{x+y}~\frac{dy}{dx} \\ \text{or,}~~ \frac{dy}{dx} \left(\frac 1y-e^{x+y}\right)=e^{x+y}-\frac 1x \\ \text{or,}~~ \frac{dy}{dx}=\frac{e^{x+y}-\frac 1x}{\frac 1y-e^{x+y}} \\ \text{or,}~~ \frac{dy}{dx}=\frac yx\cdot \frac{x \cdot e^{x+y}-1}{1-y\cdot e^{x+y}}$