Question

Question: If\(g\left( x \right) + x\sin g\left( x \right) = {x^2}\)find\(g'\left( 0 \right)\).

Short answer

Hence, the required value is \(g'\left( 0 \right) = 0\).

Step-by-step solution

 Step 1: Write the definition of implicit differentiation, the formula of the derivatives of trigonometric functions and the product rule

Implicit differentiation: Differentiate the both sides of the equation with respect to \(x\) and then solve the obtained result for \(\frac{{dy}}{{dx}}\) this process is known as Implicit differentiation.

The product rule: If \(f\) and \(g\) are differentiable functions, then \(\frac{d}{{dx}}\left( {f\left( x \right)g\left( x \right)} \right) = f\left( x \right)\frac{d}{{dx}}\left( {g\left( x \right)} \right) + g\left( x \right)\frac{d}{{dx}}\left( {f\left( x \right)} \right)\).

Find \(g'\left( 0 \right)\)by implicit differentiation

Consider the equation, \(g\left( x \right) + x\sin g\left( x \right) = {x^2}\). Differentiate both the sides of the equation w.r.t. \(x\)and simplify by using the formula of the product rule.

\(\begin{aligned}\frac{d}{{dx}}\left( {g\left( x \right) + x\sin g\left( x \right)} \right) &= \frac{d}{{dx}}\left( {{x^2}} \right)\\\frac{d}{{dx}}\left( {g\left( x \right)} \right) + \frac{d}{{dx}}\left( {x\sin g\left( x \right)} \right) &= \frac{d}{{dx}}\left( {{x^2}} \right)\\g'\left( x \right) + x\cos g\left( x \right)g'\left( x \right) + \sin g\left( x \right) \cdot 1 &= 2x\end{aligned}\)

Now consider \(x = 0\) then we have,

\(\begin{aligned}g'\left( 0 \right) + \left( 0 \right)\cos g\left( 0 \right)g'\left( 0 \right) + \sin g\left( 0 \right) \cdot 1 &= 2\left( 0 \right)\\g'\left( 0 \right) + 0 &= 0\\g'\left( 0 \right) &= 0\end{aligned}\)

Thus, \(g'\left( 0 \right) = 0\).