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What is the derivative of #f(x)=(1/x+lnx)^2#?

Answer 1

We'll need chain rule, here.

Chain rule states that #(dy)/(dx)=(dy)/(du)(du)/(dx)# and is used when it is not possible to directly derivate the function we have.

So, renaming #u=1/x+lnx#, we get:

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

#(dy)/(dx)=2(1/x+lnx)(-1/x^2+1/x)#

#(dy)/(dx)=2(-1/x^3+(1-lnx)/x^2+lnx/x)#

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Answer 2

William Abdalla

To find the derivative of ( f(x) = \left(\frac{1}{x} + \ln(x)\right)^2 ), you would use the chain rule and power rule.

The derivative is:

[ f'(x) = 2\left(\frac{1}{x} + \ln(x)\right)\left(-\frac{1}{x^2} + \frac{1}{x}\right) ]

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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.