Question

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

• A. $secx+cosx$
• B. $secx+logtanx$
• C. $(tanx)si{n}^x$
• D. None of these

Answer 1

Answer: (D)

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