Question

If $y=\left(tanx\right)^{sinx}$, then $\frac{dy}{dx}$ is equal to

• A. $sec\, x + cos\, x$
• B. $sec\, x +log\, tan \,x$
• C. $(tan\,x)sin^x$
• D. None of these

Answer 1

Answer: (D)

None of these

Answer 2

We have, $y = (tanx)^{sinx}$ Taking log on both sides, we get $logy = sinx \,log(tanx)$ Differentiating $w$.$r$.$t$. $x$, we get $\frac{1}{y} \frac{dy}{dx}=\frac{sin\,x}{tan\,x}\cdot sec^{2}x+cosx\,log\left(tanx\right)$ $=\left(tanx\right)^{sinx}\left[sec\,x+cosx\left(log\,tan\,x\right)\right]$