Answer to Question #152848 in Calculus for zeeshan wahab

Question

apply implicit differentation to determine dy/dx
a) In(sin^-1(e^x)+yx^2=1
b) tan^-1(x+y)-sin^-1(e^y+x)=4

Answer 1

a) Given that $ln(sin^{-1}(e^x))+yx^2=1$ ............(1)

Differentiating both side of (1) with respect to $x,$ we get

$\frac{d}{dx}[ln(sin^{-1}(e^x))]+\frac{d}{dx}[yx^2]=\frac{d}{dx}(1)$

$\implies \frac{1}{sin^{-1}(e^x)}.\frac{d}{dx}[sin^{-1}(e^x)]+y.\frac{d}{dx}(x^2)+x^2.\frac{d}{dx}(y)=0$

$\implies \frac{1}{sin^{-1}(e^x)}.\frac{1}{\sqrt{1-(e^x)^2}}.\frac {d}{dx}(e^x)+y.(2x)+x^2.\frac{dy}{dx}=0$

$\implies \frac{1}{sin^{-1}(e^x)}.\frac{1}{\sqrt{1-(e^x)^2}}.(e^x)+2xy+x^2.\frac{dy}{dx}=0$

$\implies x^2.\frac{dy}{dx}=-[ \frac{1}{sin^{-1}(e^x)}.\frac{1}{\sqrt{1-(e^x)^2}}.(e^x)+2xy]$

$\implies \frac{dy}{dx}=-[ \frac{1}{x^2}.\frac{1}{sin^{-1}(e^x)}.\frac{1}{\sqrt{1-(e^x)^2}}.(e^x)+\frac{2xy}{x^2}]$

$\implies \frac{dy}{dx}=-[ \frac{1}{x^2}.\frac{1}{sin^{-1}(e^x)}.\frac{1}{\sqrt{1-(e^x)^2}}.(e^x)+2(\frac{y}{x})]$

b) Given that $tan^{-1}(x+y)-sin^{-1}(e^y+x)=4$ ..........(1)

Differentiating both side of (1) with respect to $x,$ we get

$\frac{d}{dx}[tan^{-1}(x+y)]-\frac{d}{dx}[sin^{-1}(e^y+x)]=\frac{d}{dx}(4)$

$\implies \frac {1}{[1+(x+y)^2]}.\frac{d}{dx}(x+y)-\frac{1}{\sqrt{1-(e^y+x)^2}}.\frac{d}{dx}(e^y+x)=0$

$\implies \frac {1}{[1+(x+y)^2]}.(1+\frac{dy}{dx})-\frac{1}{\sqrt{1-(e^y+x)^2}}.(1+e^y\frac{dy}{dx})=0$

$\implies [\frac {1}{[1+(x+y)^2]}-\frac{e^y}{\sqrt{1-(e^y+x)^2}}].\frac{dy}{dx}+[\frac {1}{[1+(x+y)^2]}-\frac{1}{\sqrt{1-(e^y+x)^2}}]=0$

$\implies [\sqrt{1-(e^y+x)^2} -e^y[1+(x+y)^2]].\frac{dy}{dx}+[\sqrt{1-(e^y+x)^2} -[1+(x+y)^2]]=0$

$\implies \frac{dy}{dx}=-\frac{[\sqrt{1-(e^y+x)^2} -[1+(x+y)^2]]}{[\sqrt{1-(e^y+x)^2} -e^y.[1+(x+y)^2]]}$