How do you find the derivative of #2/(x+1)#?

Answer 1

This expression can be rewritten as #2(x+1)^-1#, following the exponential alw that states #a^-n=1/a^n#.

Naming #u=x+1#, we can rewrite the expression as #y=2u^-1# and, thus, derivate it according to the chain rule, which states that

#(dy)/(dx)=(dy)/(du)(du)/(dx)#

So,

#(dy)/(du)=-2u^-2#

#(du)/(dx)=1#

Thus

#(dy)/(dx)=(-2u^-2)(1)=-2(x+1)^-2=color(green)(-2/(x+1)^2)#

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Answer 2

Scarlett Allison

To find the derivative of ( \frac{2}{x+1} ), you can use the quotient rule, which states that if you have a function ( \frac{f(x)}{g(x)} ), its derivative is ( \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} ). Applying this rule, you would find the derivative of the numerator and denominator separately, then use the quotient rule to combine them.

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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.