# Differentiate the functions with respect to x: $cos \; x^3. sin^2 (x)^5$

$\begin{array}{1 1} 10sin^{19}xcosx-3x^2cosx^2sinx^3 \\ 10sin^{9}xcosx-3x^2cosx^3sinx^3 \\ 10sin^{19}xcosx-3x^2cosx^3sinx^3 \\ 10sin^{19}xcosx-3xcosx^3sinx^3 \end{array}$

Toolbox:
• According to the Product Rule for differentiation, given two functions $u$ and $v, \;\large \frac {d(uv)}{dx}$$= u \large\frac{dv}{dx}$$+ v\large \frac{du}{dx}$
• According to the Chain Rule for differentiation, given two functions $f(x)$ and $g(x)$, and $y=f(g(x)) \rightarrow y' = f'(g(x)).g'(x)$.
• $\; \large \frac{d(sin^2x)}{dx} $$= sin 2x • \; \large \frac{d(cosx)}{dx}$$=-sinx$
Given $y = cos \; x^3. sin^2 (x)^5,\;$ we can apply the product rule of differentiation, as this is of the form $y = uv$
$\textbf{Step 1}$:
Let $u = cos \; x^3$. This is of the form $u = f(g(x)$, where $g(x) = x^3$.
We can apply the Chain rule. According to the Chain Rule for differentiation, given two functions $f(x)$ and $g(x)$, and $y=f(g(x)) \rightarrow y' = f'(g(x)).g'(x)$.
$\Rightarrow g'(x) = 3\;x^2$
$\; \large \frac{d(cosx)}{dx} $$=-sinx \Rightarrow f'(g(x)) = -sin(g(x)) = -sin(x^3) \Rightarrow u' = f'(g(x)).g'(x) =- 3\;x^2.sin(x^3) \textbf{Step 2}: Let v = sin^2 (x)^5. We can rewrite this as v = sin^{10}x This is of the form v = f(g(x), where g(x) = sinx We can apply the Chain rule. According to the Chain Rule for differentiation, given two functions f(x) and g(x), and y=f(g(x)) \rightarrow y' = f'(g(x)).g'(x). \Rightarrow g'(x) = cosx \Rightarrow f'(g(x)) = 10\;sin^9x \Rightarrow u' = f'(g(x)).g'(x) = 10\;sin^9x\;cosx \textbf{Step 3}: According to the Product Rule for differentiation, given two functions u and v, \;\large \frac {d(uv)}{dx}$$ = u \large\frac{dv}{dx} $$+ v\large \frac{du}{dx} \Rightarrow \large \frac {d(uv)}{dx} =$$ cos\;x^3. -3 x^2.sin\;x^3 + sin^{10}x.10\;sin^9x\;cosx$
$\Rightarrow \large \frac {d(uv)}{dx}$$= 10sin^{19}xcosx-3x^2cosx^3sinx^3$