How do you differentiate #g(x) = sqrt(x-3)e^(4x)cos(7x)# using the product rule?

Answer 1

#d/dx f(x) = (1-12e^(4x))/(2sqrt(x-3e^(4x)))(cos(7x))+(-7sin(7x))(sqrt(x-3e^(4x)))#

I'm assuming your original post meant #sqrt(x-3e^(4x))cos(7x)# and this was a simple typo in inferring this question. Therefore let us begin by recalling the product rule.
#(fg)'=f'g +fg'# where #f and g# are differentiable.
Let us define #f and g#
#f(x) = sqrt(x-3e^(4x))# #g(x)=cos(7x)#

Let us break down this problem into manageable chunks by differentiating each function independently and entering our solution at the end of the problem.

Therefore, #d/dx f(x)# requires us to use the chain rule.
where our inner function is #x^(1/2)# and our outer function is #(x-3e^(4x))#
recall #(f circ g)' = (f' circ g)* g'#
#f' = (1/2)(x^(-1/2))# or #(1)/(2sqrt(x))# of our inner function where we must plug in #g = (x-3e^(4x))# #(1)/(2sqrt((x-3e^(4x)))# now find #g'#
#g = (x-3e^(4x))#
#g' = 1-12e^(4x)# recalling that differentiation splits across sums and differences making two derivatives. Noting #d/dx e^x = (x')e^x#
Where our #x=(4x)#
Furthermore, #d/dx f(x) = (1-12e^(4x))/(2sqrt(x-3e^(4x))#
Now let us find #d/dx g(x) = cos(7x)# recalling that we must use the chain rule again.
#d/dx g(x) = -7sin(7x)# where
#f = cos(x) and g=7x# #(f circ g)' = (f' circ g)* g'#
#d/dx cos(x) = -sin(x)#
#d/dx 7x = 7#
Recall our original problem #sqrt(x-3e^(4x))cos(7x)# #(fg)'=f'g +fg'# where #f and g# are differentiable.
#f = sqrt(x-3e^(4x)) and g=cos(7x)# also
#f' = (1-12e^(4x))/(2sqrt(x-3e^(4x))# and #g' =-7sin(7x)#
#(1-12e^(4x))/(2sqrt(x-3e^(4x)))(cos(7x))+(-7sin(7x))(sqrt(x-3e^(4x)))#
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Answer 2

To differentiate ( g(x) = \sqrt{x-3} e^{4x} \cos(7x) ) using the product rule, follow these steps:

  1. Identify the functions being multiplied together: ( f(x) = \sqrt{x-3} e^{4x} ) and ( h(x) = \cos(7x) ).
  2. Apply the product rule: ( g'(x) = f'(x)h(x) + f(x)h'(x) ).
  3. Find the derivatives of ( f(x) ) and ( h(x) ):
    • ( f'(x) = \frac{1}{2\sqrt{x-3}} e^{4x} + \sqrt{x-3} \cdot 4e^{4x} ).
    • ( h'(x) = -7\sin(7x) ).
  4. Plug the derivatives into the product rule formula and simplify.

The derivative of ( g(x) ) using the product rule is:

[ g'(x) = \left(\frac{1}{2\sqrt{x-3}} e^{4x} + \sqrt{x-3} \cdot 4e^{4x}\right) \cos(7x) + \sqrt{x-3} e^{4x} \left(-7\sin(7x)\right) ]

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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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