If #f(x)= cos5 x # and #g(x) = e^(3+4x ) #, how do you differentiate #f(g(x)) # using the chain rule?
Leibniz's notation can come in handy.
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To differentiate ( f(g(x)) ) using the chain rule, follow these steps:
- Find the derivative of the outer function with respect to the inner function.
- Find the derivative of the inner function with respect to ( x ).
- Multiply the results from steps 1 and 2 together.
Given ( f(x) = \cos^5(x) ) and ( g(x) = e^{3+4x} ), we proceed as follows:
-
Find ( f'(u) ), where ( u = g(x) ): [ f'(u) = 5\cos^4(u)(-\sin(u)) ]
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Find ( g'(x) ): [ g'(x) = 4e^{3+4x} ]
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Multiply ( f'(u) ) and ( g'(x) ): [ f'(g(x)) \cdot g'(x) = 5\cos^4(g(x))(-\sin(g(x))) \cdot 4e^{3+4x} ]
So, the derivative of ( f(g(x)) ) with respect to ( x ) using the chain rule is: [ -20\cos^4(g(x))\sin(g(x))e^{3+4x} ]
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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.
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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