If #f(x)= sin6 x # and #g(x) = e^(3+2x ) #, how do you differentiate #f(g(x)) # using the chain rule?

Question

If #f(x)= sin6  x # and #g(x)  = e^(3+2x   )  #, how do you differentiate #f(g(x))  # using the chain rule?

Luke Alvarez · Accepted Answer

Step by step explanation is given below.

Chain rule #(f(g(x)))' = f'(g(x))**g'(x)# #f(x) = sin(6x)# Differentiating with respect to x using chain rule #f'(x) = cos(6x)d/dx(6x)# #f'(x) = 6cos(6x)# #g(x) = e^(3+2x)# Differentiating with respect to x using chain rule #g'(x) = e^(3+2x)d/dx(3+2x)# #g'(x)=e^(3+2x)(2)# #g'(x)=2e^(3+2x)# We are to find derivative of #f(g(x))# #f'(g(x)) = 6cos(6e^(3+2x))# Using the chain rule #(f(g(x)))' = f'(g(x))**g'(x)# #(f(g(x)))' = 6cos(6e^(3+2x))(2e^(3+2x))# #(f(g(x)))' = 12e^(3+2x)cos(6e^(3+2x))# Answer

Paisley Johnson · Answer

undefined To differentiate $f(g(x))$ using the chain rule, where $f(x) = \sin^6(x)$ and $g(x) = e^{3+2x}$, we apply the chain rule formula:

$$
\frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x)
$$

First, differentiate $f(x) = \sin^6(x)$ with respect to $x$:

$$
f'(x) = 6\sin^5(x) \cos(x)
$$

Then, differentiate $g(x) = e^{3+2x}$ with respect to $x$:

$$
g'(x) = (3+2x)' e^{3+2x} = 2e^{3+2x}
$$

Next, substitute $g(x)$ into $f'(x)$:

$$
f'(g(x)) = 6\sin^5(g(x)) \cos(g(x)) = 6\sin^5(e^{3+2x}) \cos(e^{3+2x})
$$

Finally, apply the chain rule $f'(g(x)) \cdot g'(x)$:

$$
\frac{d}{dx}f(g(x)) = (6\sin^5(e^{3+2x}) \cos(e^{3+2x})) \cdot (2e^{3+2x})
$$

$$
= 12e^{3+2x} \sin^5(e^{3+2x}) \cos(e^{3+2x})
$$

Thus, the derivative of $f(g(x))$ with respect to $x$ is $12e^{3+2x} \sin^5(e^{3+2x}) \cos(e^{3+2x})$.