If #f(x)= cos5 x # and #g(x) = e^(3+4x ) #, how do you differentiate #f(g(x)) # using the chain rule?

Answer 1

Leibniz's notation can come in handy.

#f(x)=cos(5x)#
Let #g(x)=u#. Then the derivative:
#(f(g(x)))'=(f(u))'=(df(u))/dx=(df(u))/(dx)(du)/(du)=(df(u))/(du)(du)/(dx)=#
#=(dcos(5u))/(du)*(d(e^(3+4x)))/(dx)=#
#=-sin(5u)*(d(5u))/(du)*e^(3+4x)(d(3+4x))/(dx)=#
#=-sin(5u)*5*e^(3+4x)*4=#
#=-20sin(5u)*e^(3+4x)#
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Answer 2

To differentiate ( f(g(x)) ) using the chain rule, follow these steps:

  1. Find the derivative of the outer function with respect to the inner function.
  2. Find the derivative of the inner function with respect to ( x ).
  3. 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:

  1. Find ( f'(u) ), where ( u = g(x) ): [ f'(u) = 5\cos^4(u)(-\sin(u)) ]

  2. Find ( g'(x) ): [ g'(x) = 4e^{3+4x} ]

  3. 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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Answer from HIX Tutor

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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