Question

If Xy = Ex-y, prove that . dy/DX =logx/(log ec)2​

Answer 1

$\\ \underline \red{\large\rm \underline{Given :- }}$

• $\sf{x}^{y} = {e}^{x - y}$

$\\ \underline \red{\large\rm \underline{To \: prove :- }}$

• $\sf \dfrac{dy}{dx} = \dfrac{ \log \: x}{ {{(1}\log \: x)^{2} }}$

$\\ \underline \red{\large\rm \underline{ \: Proof :- }}$

$\\ \: \: \sf{x}^{y} = {e}^{x - y}$

To prove this , we need to take log in both side of above equation ,

So,

$\\ \: \sf\log \: \sf{x}^{y} = \log \: {e}^{x - y}$

Therefore,

$\\ \sf \: \: \: \: y \: log \: x \: = (x - y) \: log \: e$

$\\ \sf \: \: \: \: y \: log \: x \: = (x - y) \:$

$\\ \sf \: \: \: \: \: log \: x \: = \dfrac{ (x - y) }{y}\:$

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \qquad \: \qquad Or$

$\\ \: \: \therefore \: \sf \: \: \: \: \: \: \frac{y}{x} \: = \dfrac{ 1 }{(1 + \log \: x)}\: \dots \: (1)$

Now, let's Differentiating bothe side with respect to x,

$\\ \sf \: \: y \frac{1}{x} + \log \: x \: \times \: \frac{dy}{dx} = 1 - \frac{dy}{dx}$

$\\ \sf \: \: \: \: : \longmapsto \: \log \: x \: \times \: \frac{dy}{dx} + \frac{dy}{dx} = 1 - \frac{y}{x}$

$\\ \sf \: \: \: \: : \longmapsto \:(1 + \log \: x )\: \frac{dy}{dx} = 1 - \frac{y}{x}$

$\\ \sf \: \: \: \: : \longmapsto \:(1 + \log \: x )\: \frac{dy}{dx} = \frac{1( x)}{x} - \frac{y}{x}$

$\\ \sf \: \: \: \: : \longmapsto \:(1 + \log \: x )\: \frac{dy}{dx} = \frac{y \log \: x}{x}$

$\\ \: \qquad \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \dots \sf \:by (1) \bigg \{\: \sf \: \: \frac{y}{x} \: = \dfrac{ 1 }{(1 + \log \: x)}\: \bigg \}$

$\\ \sf \: \: \: \: \:\: \frac{dy}{dx} = \frac{y \log \: x}{x.(1 + \log \: x)}$

$\\ \: \qquad \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \dots \sf \:by (1) \bigg \{\: \sf \: \: \frac{y}{x} \: = \dfrac{ 1 }{(1 + \log \: x)}\: \bigg \}$

$\\ \sf \: \: \: \: \: \: \qquad \boxed{\sf \dfrac{dy}{dx} = \dfrac{ \log \: x}{ {{(1}\log \: x)^{2} }} }$

Hence proved.