How do you differentiate #f(x)=(2cosx)/(x+1)#?

Answer 1

#(-2(x+1)sinx-2cosx)/(x+1)^2#

differentiate using the #color(blue)"quotient rule"#

If #f(x)=(g(x))/(h(x))# then

#color(red)(bar(ul(|color(white)(a/a)color(black)(f'(x)=(h(x)g'(x)-g(x)h'(x))/(h(x))^2)color(white)(a/a)|)))# #color(blue)"----------------------------------------------------------------"#

here #g(x)=2cosxrArrg'(x)=-2sinx#

and #h(x)=x+1rArrh'(x)=1# #color(blue)"-----------------------------------------------------------------"# substitute these values into f'(x)

#rArrf'(x)=((x+1)(-2sinx)-2cosx(1))/(x+1)^2#

#=(-2(x+1)sinx-2cosx)/(x+1)^2#

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Answer 2

Emily Ammerman

To differentiate the function ( f(x) = \frac{2\cos(x)}{x+1} ), we use the quotient rule.

[ f'(x) = \frac{(x+1)(-2\sin(x)) - (2\cos(x))(1)}{(x+1)^2} ]

[ f'(x) = \frac{-2(x+1)\sin(x) - 2\cos(x)}{(x+1)^2} ]

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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.