Question

$\frac{d}{dx}\log \left[\tan \right\{{\tan }^{-1}\left(\sin hx\right)\left\}\right]=$____, where $h=\text{constant}.$

• A. $h\sin h\left(2x\right)$
• B. $h\cot h(x$)
• C. $h^2\cot h\left(2x\right)$
• D. $-h\sin h\left(2x\right)$

Answer 1

The correct option is B $h\cot h(x$)

Use the inverse trigonometry identities: $\tan \left({\tan }^{-1}x\right)=x$

So, $\log \left(\tan \right({\tan }^{-1}\left(\sin \right(hx\left)\right))=\log (\sin \left(hx\right))$

Now, apply the chain rule, $\frac{df}{dx}=\frac{df}{du}\times \frac{du}{dv}\times \frac{dv}{dx}$

Therefore, $\frac{d}{dx}\left(\log \right(\sin \left(hx\right)\left)\right)=\left(\log \right(\sin \left(hx\right)){)}^{\prime }\times (\sin \left(hx\right){)}^{\prime }\times (hx{)}^{\prime }$

$=\frac{1}{\sin \left(hx\right)}\times \cos \left(hx\right)\times h$

$=h\cot \left(hx\right)$