Question

If $\left(f\right(x){)}^{g\left(y\right)}=e^{f\left(x\right)-g\left(y\right)}$ then $\frac{dy}{dx}=$.

• A. $\frac{f^1\left(x\right)\log f\left(x\right)}{g^1\left(y\right)(1+\log f(x){)}^2}$
• B. $\frac{f^1\left(x\right)\log f\left(x\right)}{g^1\left(y\right)(1+\log f(x){)}^3}$
• C. $\frac{f^1\left(x\right).\log f\left(x\right)}{g^{-1}\left(y\right)(1-\log f(x){)}^2}$
• D. $\frac{f^1\left(x\right)\log f\left(x\right)}{g\left(y\right)(1+\log f(x){)}^2}$

Answer 1

Answer: (A)

$\frac{f^1\left(x\right)\log f\left(x\right)}{g^1\left(y\right)(1+\log f(x){)}^2}$

${f\left(x\right)}^{g\left(y\right)}=e^{f\left(x\right)-g\left(y\right)}$

$\log {f\left(x\right)}^{g\left(y\right)}=f\left(x\right)-g\left(y\right)$

$g\left(y\right)\log f\left(x\right)=f\left(x\right)-g\left(y\right)$

$f\left(x\right)=g\left(y\right)[1+\log f\left(x\right)]$ ...(1)

$f^1\left(x\right)=g\left(y\right)\left[\frac{f^1\left(x\right)}{f\left(x\right)}\right]+g^1\left(y\right)\times [1+\log f\left(x\right)]\times \frac{dy}{dx}$

Rearranging,

$f^1\left(x\right)\left[\frac{f\left(x\right)-g\left(y\right)}{f\left(x\right)}\right]=g^1\left(y\right)\times [1+\log f\left(x\right)]\times \frac{dy}{dx}$

from eqn. (1)

$f^1\left(x\right)\left[\frac{g\left(y\right)\times \log f\left(x\right)}{f\left(x\right)}\right]=g^1\left(y\right)\times [1+\log f\left(x\right)]\times \frac{dy}{dx}$

again, from eqn (1),

$f^1\left(x\right)\left[\frac{\log f\left(x\right)}{[1+\log f\left(x\right)]}\right]=g^1\left(y\right)\times [1+\log f\left(x\right)]\times \frac{dy}{dx}$

$\frac{dy}{dx}=\frac{f^1\left(x\right)}{g^1\left(y\right)}\times \left[\frac{\log f\left(x\right)}{{[1+\log f\left(x\right)]}^2}\right]$