# How do you calculate the derivate for #f(x)=[(x+2)/(x+1)](2x-7)#?

Using a combination of Product Rule and Quotient Rule.

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To calculate the derivative of ( f(x) = \frac{{(x+2)}}{{(x+1)}}(2x-7) ), you can use the product rule, which states that the derivative of the product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function.

Applying the product rule to ( f(x) ), we get:

( f'(x) = \left(\frac{{d}}{{dx}}\left(\frac{{x+2}}{{x+1}}\right)\right)(2x-7) + \left(\frac{{x+2}}{{x+1}}\right)\left(\frac{{d}}{{dx}}(2x-7)\right) )

Now, differentiate each part separately:

( \frac{{d}}{{dx}}\left(\frac{{x+2}}{{x+1}}\right) = \frac{{(x+1)(1) - (x+2)(1)}}{{(x+1)^2}} )

( \frac{{d}}{{dx}}(2x-7) = 2 )

Plug these derivatives back into the equation:

( f'(x) = \frac{{(x+1) - (x+2)}}{{(x+1)^2}}(2x-7) + \frac{{x+2}}{{x+1}}(2) )

Now, simplify:

( f'(x) = \frac{{-1}}{{(x+1)^2}}(2x-7) + 2\frac{{x+2}}{{x+1}} )

( f'(x) = \frac{{-2x+7}}{{(x+1)^2}} + 2\frac{{x+2}}{{x+1}} )

( f'(x) = \frac{{-2x+7}}{{(x+1)^2}} + \frac{{2(x+2)}}{{x+1}} )

( f'(x) = \frac{{-2x+7+2(x+2)}}{{(x+1)^2}} )

( f'(x) = \frac{{-2x+7+2x+4}}{{(x+1)^2}} )

( f'(x) = \frac{{11}}{{(x+1)^2}} )

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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.

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