How do you use the chain rule to differentiate #f(x)=((3x-5)/(4x-7))^-6#?

Answer 1

#f'(x) = frac(6(4 x - 7)^(5))((3 x - 5)^(7))#

We have: #f(x) = (frac(3 x - 5)(4 x - 7))^(- 6)#

This function can be differentiated using a combination of the "quotient rule" and the "chain rule".

Let #u = frac(3 x - 5)(4 x - 7)# and #v = u^(- 6)#:
#Rightarrow f'(x) = u' cdot v'#
#Rightarrow f'(x) = (frac((4 x - 7) cdot frac(d)(dx)(3 x - 5) - (3 x - 5) cdot frac(d)(dx)(4 x - 7))((4 x - 7)^(2))) cdot (- 6 u^(- 7))#
#Rightarrow f'(x) = (frac((4 x - 7)(3) - (3 x - 5)(4))((4 x - 7)^(2))) cdot (- 6 u^(- 7)#
#Rightarrow f'(x) = (frac(12 x - 21 - 12 x + 20)((4 x - 7)^(2))) cdot (- 6 u^(- 7))#
#Rightarrow f'(x) = (frac(- 1)((4 x - 7)^(2))) cdot (- 6 u^(- 7))#
#Rightarrow f'(x) = frac(6 u^(- 7))((4 x - 7)^(2))#
Let's replace #u# with #frac(3 x - 5)(4 x - 7)#:
#Rightarrow f'(x) = frac(6(frac(3 x - 5)(4 x - 7))^(- 7))((4 x -7)^(2))#
#Rightarrow f'(x) = frac(6(frac((3 x - 5)^(- 7))((4 x - 7)^(- 7))))((4 x -7)^(2))#
#Rightarrow f'(x) = frac(6(3 x - 5)^(- 7))((4 x - 7)^(- 5))#
#Rightarrow f'(x) = frac(frac(6)((3 x - 5)^(7)))(frac(1)((4 x - 7)^(5)))#
#Rightarrow f'(x) = frac(6(4 x - 7)^(5))((3 x - 5)^(7))#
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Answer 2

To differentiate ( f(x) = \left( \frac{3x - 5}{4x - 7} \right)^{-6} ) using the chain rule, follow these steps:

  1. Rewrite the function as ( f(x) = \left( g(x) \right)^{-6} ), where ( g(x) = \frac{3x - 5}{4x - 7} ).

  2. Find the derivative of the inner function ( g(x) ) with respect to ( x ), denoted as ( g'(x) ).

  3. Apply the chain rule, which states that if ( y = u^n ), where ( u ) is a differentiable function of ( x ), then ( \frac{dy}{dx} = n \cdot u^{n-1} \cdot \frac{du}{dx} ).

  4. Substitute ( g(x) ) and ( g'(x) ) into the chain rule formula to find ( \frac{df}{dx} ), the derivative of ( f(x) ) with respect to ( x ).

So, ( \frac{df}{dx} = -6 \cdot \left( \frac{3x - 5}{4x - 7} \right)^{-7} \cdot \frac{d}{dx} \left( \frac{3x - 5}{4x - 7} \right) ).

To find ( \frac{d}{dx} \left( \frac{3x - 5}{4x - 7} \right) ), use the quotient rule: ( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} ), where ( u = 3x - 5 ) and ( v = 4x - 7 ).

After calculating ( \frac{d}{dx} \left( \frac{3x - 5}{4x - 7} \right) ), substitute the result back into ( \frac{df}{dx} ) to get the final answer.

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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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