4.) It is given that e^xy=x[(x+1)^3] / (x^2+1) ,where x>0
determine the equation of the tangent line at x=1
\(e^xy=\frac{x(x+1)^3}{(x^2+1)}\\ \mbox{Left hand side}\\ \frac{d}{dx}e^xy=e^xy+e^x\frac{dy}{dx}\\ \mbox{Right hand side}\\ Let\;\;u=x(x+1)^3\qquad \qquad \qquad \qquad and \qquad v=x^2+1\\ \qquad \;\;u'=(x+1)^3+3x (x+1)^2 \qquad and \qquad v'=2x\\ \frac{d}{dx}\frac{x(x+1)^3}{(x^2+1)}=\frac{(x^2+1)[(x+1)^3+3x (x+1)^2 ]-[x(x+1)^3*2x]}{(x^2+1)^2}\\ \frac{d}{dx}\frac{x(x+1)^3}{(x^2+1)}=\frac{(x^2+1)[(x+1)^3+3x (x+1)^2 ]-[2x^2(x+1)^3]}{(x^2+1)^2}\\so\\ e^xy+e^x\frac{dy}{dx}=\frac{(x^2+1)[(x+1)^3+3x (x+1)^2 ]-[2x^2(x+1)^3]}{(x^2+1)^2}\\ \)
\(e^xy+e^x\frac{dy}{dx}=\frac{(x^2+1)[(x+1)^3+3x (x+1)^2 ]-[2x^2(x+1)^3]}{(x^2+1)^2}\\ when\;\; x=1\\ ey+e\frac{dy}{dx}=\frac{2*[8+3*4]-[2*8]}{4}\\ ey+e\frac{dy}{dx}=\frac{40-16}{4}\\ ey+e\frac{dy}{dx}=6\\ \frac{dy}{dx}=\frac{6-ey}{e}\\\)
NOW
\( e^xy=\frac{x[(x+1)^3] }{ (x^2+1)}\\ When\;\;x=1\\ ey=\frac{8}{ 2}\\ y=\frac{4}{ e}\\ so\;\;when\;\;x=1\\ \frac{dy}{dx}=\frac{6-e*\frac{4}{e}}{e}\\ \frac{dy}{dx}=\frac{2}{e}\qquad \mbox{This is the gradient ofthe tangent at x=1}\\\)
So now for the equation of the tangent at (1, 4/e)
\(\frac{2}{e}=\frac{y-\frac{4}{e}}{x-1}\\ \frac{2(x-1)}{e}=\frac{ey-4}{e}\\ 2(x-1)=ey-4\\ 2x-2+4=ey\\ y=\frac{2x+2}{e}\\\)
I worked this out and coded it at the same time so it could be riddled with errors or it could just be plain wrong.
OR maybe there is a much quicker way..... idk .... But that is my shot :)
