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

Step by step explanation is given below.

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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})).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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