How do you differentiate #f(x)= -1 / (2x-7 )# using the quotient rule?

Answer 1

#frac{d}{dx}(-frac{1}{2x-7})=frac{2}{(2x-7)^2}#

#frac{d}{dx}(-frac{1}{2x-7})#

Eliminating the constant,

#(acdot f)^'=acdot f^'#
#=-frac{d}{dx}(frac{1}{2x-7})#
#=-frac{d}{dx}((2x-7)^{-1})#

Using the chain rule

#frac{df(u)}{dx}=frac{df}{du}cdot frac{du}{dx}#
#Let,2x-7=u#
#=-frac{d}{du}(u^{-1})frac{d}{dx}(2x-7)#

WE know,

#frac{d}{du}(u^{-1})=-frac{1}{u^2}#
#frac{d}{dx}(2x-7)=2#
#=-(-frac{1}{u^2})2#
Substitute back #=-(-frac{1}{u^2})# #=-(-frac{1}{(2x-7)^2})2#
#=-(-frac{1}{(2x-7)^2})2#
Simplify, #frac{2}{(2x-7)^2}#
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Answer 2

To differentiate ( f(x) = \frac{{-1}}{{2x - 7}} ) using the quotient rule, you would apply the formula ( \frac{{d}}{{dx}}\left(\frac{{f(x)}}{{g(x)}}\right) = \frac{{f'(x)g(x) - f(x)g'(x)}}{{[g(x)]^2}} ), where ( f(x) = -1 ) and ( g(x) = 2x - 7 ). Then, differentiate ( f(x) ) and ( g(x) ) separately and plug the results into the quotient rule formula.

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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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