# How do you find the derivative of  y= a^3 + cos^3 x ?

Jun 25, 2016

$- 3 \sin x {\cos}^{2} x$

#### Explanation:

Here the variable is x and a is a constant.

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder:"d/dx"(constant)}} = 0$

Express ${\cos}^{3} x = {\left(\cos x\right)}^{3}$

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

$\frac{d}{\mathrm{dx}} \left(f \left(g \left(x\right)\right)\right) = f ' \left(g \left(x\right)\right) . g ' \left(x\right) \ldots \ldots . . \left(A\right)$
$\text{----------------------------------------------------}$
$f \left(g \left(x\right)\right) = {\left(\cos x\right)}^{3} \Rightarrow f ' \left(g \left(x\right)\right) = 3 {\left(\cos x\right)}^{2} = 3 {\cos}^{2} x$

$g \left(x\right) = \cos x \Rightarrow g ' \left(x\right) = - \sin x$
$\text{----------------------------------------------------}$
Substitute these values into(A)

$\Rightarrow 3 {\cos}^{2} x \left(- \sin x\right) = - 3 \sin x {\cos}^{2} x$

So y=${a}^{3} + {\cos}^{3} x$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = 0 - 3 \sin x {\cos}^{2} x = - 3 \sin x {\cos}^{2} x$