Question

Find the derivative of the function.

41. \(y = {\rm{si}}{{\rm{n}}^2}\left( {{x^2} + 1} \right)\)

Short answer

The derivative of the function is \(4x{\rm{sin}}\left( {{x^2} + 1} \right){\rm{cos}}\left( {{x^2} + 1} \right)\), that is, \(y' = 4x{\rm{sin}}\left( {{x^2} + 1} \right){\rm{cos}}\left( {{x^2} + 1} \right)\).

Step-by-step solution

Use Power and Chain rule

The power rule of differentiation implies that \(\frac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\), where \(n\) is any real number.

According to the chain rule, \(F'\left( x \right) = f'\left( {g\left( x \right)} \right) \cdot g'\left( x \right)\), where \(F\left( x \right) = f\left( {g\left( x \right)} \right)\) and \(f\) is differentiable at \(g\left( x \right)\) .

Find the derivative of \(y\) with respect to \(x\).

\(\begin{aligned}y' &= \frac{d}{{dx}}\left( {{\rm{si}}{{\rm{n}}^2}\left( {{x^2} + 1} \right)} \right)\\ &= 2{\rm{sin}}\left( {{x^2} + 1} \right)\frac{d}{{dx}}\left( {{\rm{sin}}\left( {{x^2} + 1} \right)} \right)\end{aligned}\)

Chain rule and sum rule of differentiation

Use sum rule of differentiation which implies that \(\frac{d}{{dx}}\left( {f\left( x \right) + g\left( x \right)} \right) = \frac{d}{{dx}}f\left( x \right) + \frac{d}{{dx}}g\left( x \right)\).

Apply chain and sum rule of differentiation.

\(\begin{aligned}y' &= 2{\rm{sin}}\left( {{x^2} + 1} \right)\frac{d}{{dx}}\left( {{\rm{sin}}\left( {{x^2} + 1} \right)} \right)\\ &= 2{\rm{sin}}\left( {{x^2} + 1} \right){\rm{cos}}\left( {{x^2} + 1} \right)\frac{d}{{dx}}\left( {{x^2} + 1} \right)\\ &= 2{\rm{sin}}\left( {{x^2} + 1} \right){\rm{cos}}\left( {{x^2} + 1} \right)\left( {\frac{d}{{dx}}\left( {{x^2}} \right) + \frac{d}{{dx}}\left( 1 \right)} \right)\\ &= 2{\rm{sin}}\left( {{x^2} + 1} \right){\rm{cos}}\left( {{x^2} + 1} \right)\left( {2x + 0} \right)\\ &= 4x{\rm{sin}}\left( {{x^2} + 1} \right){\rm{cos}}\left( {{x^2} + 1} \right)\end{aligned}\)

Thus, the derivative of the function is \(4x{\rm{sin}}\left( {{x^2} + 1} \right){\rm{cos}}\left( {{x^2} + 1} \right)\), that is, \(y' = 4x{\rm{sin}}\left( {{x^2} + 1} \right){\rm{cos}}\left( {{x^2} + 1} \right)\).