Question

Find the derivatives of each of the functions in Exercises 17–50. In some cases it may be convenient to do some preliminary algebra

$f\left(x\right)=\sin \left({\sin }^{-1}x\right)$.

Short answer

The derivative is$1.$

Step-by-step solution

Step 1. Given Information.

The given function is$f\left(x\right)=\sin \left({\sin }^{-1}x\right)$.

Step 2. Preliminary Algebra.

We know,

$$\begin{gathered} \frac{d}{dx}\left(\sin x\right)=\cos x \\ \frac{d}{dx}\left({\sin }^{-1}x\right)=\frac{1}{\sqrt{1-x^2}} \\ \left(fg{)}^{\text{'}}\right(x)=f^{\text{'}}(x\left)g\right(x)+f(x\left)g^{\text{'}}\right(x) \end{gathered}$$

Step 3. Derivative of the function.

The derivative of the function will be,

$$\begin{gathered} \frac{d}{dx}\left(\sin \left({\sin }^{-1}x\right)\right)=\frac{d}{dx}\left(x\right) \\ Since,\left[{\sin }^{-1}x\text{is the inverse of}\sin x\right] \\ Therefore, \\ \frac{d}{dx}\left(\sin \left({\sin }^{-1}x\right)\right)=1 \end{gathered}$$