# If f(x) =sinx  and g(x) = (x+3)^3 , what is f'(g(x)) ?

Apr 4, 2018

#### Explanation:

Hmm...

Let's let $g \left(x\right) = y$

We now apply this to $f \left(x\right)$.

$\implies f \left(y\right) = \sin \left(y\right)$ Substitute $y$ with ${\left(x + 3\right)}^{3}$

$\implies f \left(g \left(x\right)\right) = \sin \left({\left(x + 3\right)}^{3}\right)$

If you meant to find just $f ' \left(g \left(x\right)\right)$...

Just find the derivative of $\sin \left(y\right)$

Since derivative of $\sin \left(x\right)$ is $\cos \left(x\right)$...

$\implies f ' \left(g \left(x\right)\right) = \cos \left({\left(x + 3\right)}^{3}\right)$

If you meant to ask for $\frac{d}{\mathrm{dx}} \left[f \left(g \left(x\right)\right)\right] ,$ you use the chain rule:

$\frac{d}{\mathrm{dx}} \left[f \left(g \left(x\right)\right)\right] = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$

We already know that $f ' \left(g \left(x\right)\right) = \cos \left({\left(x + 3\right)}^{3}\right)$

$\implies \frac{d}{\mathrm{dx}} \left[f \left(g \left(x\right)\right)\right] = \cos \left({\left(x + 3\right)}^{3}\right) \cdot \frac{d}{\mathrm{dx}} \left[{\left(x + 3\right)}^{3}\right]$

We use the power rule:

$\frac{d}{\mathrm{dx}} \left[{x}^{n}\right] = n {x}^{n - 1}$ if $n$ is a constant.

$\implies \frac{d}{\mathrm{dx}} \left[f \left(g \left(x\right)\right)\right] = \cos \left({\left(x + 3\right)}^{3}\right) \cdot 3 {\left(x + 3\right)}^{3 - 1} \cdot \frac{d}{\mathrm{dx}} \left[{x}^{1} + 3 {x}^{0}\right]$

Note here that the chain rule is applied again.

$\implies \frac{d}{\mathrm{dx}} \left[f \left(g \left(x\right)\right)\right] = \cos \left({\left(x + 3\right)}^{3}\right) \cdot 3 {\left(x + 3\right)}^{2} \cdot 1 \cdot {x}^{1 - 1} + 3 \cdot 0 \cdot {x}^{0 - 1}$

$\implies \frac{d}{\mathrm{dx}} \left[f \left(g \left(x\right)\right)\right] = \cos \left({\left(x + 3\right)}^{3}\right) \cdot 3 {\left(x + 3\right)}^{2} \cdot 1 \cdot 1 + 0$

$\implies \frac{d}{\mathrm{dx}} \left[f \left(g \left(x\right)\right)\right] = \cos \left({\left(x + 3\right)}^{3}\right) \cdot 3 {\left(x + 3\right)}^{2}$

That is the entire derivative calculated.