# What is the derivative of this function sin(x^2+1)?

Jan 23, 2017

$2 x \cos \left({x}^{2} + 1\right)$

#### Explanation:

differentiate using the $\textcolor{b l u e}{\text{chain rule}}$

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder }} \textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{\frac{d}{\mathrm{dx}} \left[\sin \left(f \left(x\right)\right)\right] = \cos \left(f \left(x\right)\right) . f ' \left(x\right)} \textcolor{w h i t e}{22} |}}}$

$\Rightarrow \frac{d}{\mathrm{dx}} \left[\sin \left({x}^{2} + 1\right)\right] = \cos \left({x}^{2} + 1\right) . \frac{d}{\mathrm{dx}} \left({x}^{2} + 1\right)$

$= 2 x \cos \left({x}^{2} + 1\right)$

Jan 23, 2017

$2 x \cos \left({x}^{2} + 1\right)$.

#### Explanation:

Let $y = \sin \left({x}^{2} + 1\right) , \text{ and, let } {x}^{2} + 1 = t$.

$\therefore y = \sin t , \mathmr{and} , t = {x}^{2} + 1$.

Thus, $y$ is a function of $t , \text{ and, "t" of } x$.

As per the Chain Rule , then, we have,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dt}} \frac{\mathrm{dt}}{\mathrm{dx}} \ldots \ldots \ldots \ldots . \left(\ast\right)$

Now, $y = \sin t \Rightarrow \frac{\mathrm{dy}}{\mathrm{dt}} = \cos t \ldots \ldots \ldots \ldots \ldots \left(1\right) , \text{ and, }$

$t = {x}^{2} + 1 \Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}} = 2 x \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \left(2\right)$.

From $\left(\ast\right) , \left(1\right) \mathmr{and} \left(2\right)$, we have,

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 x \cos t = 2 x \cos \left({x}^{2} + 1\right)$.

Enjoy Maths., and, spread the Joy!