Question

Differentiate with respect to x
1/ln(x)

Answer 1

\[f(x)=\frac{1}{ln(x)}\]\[f(x)=[ln(x)]^{-1}\]\[f'(x)=-\frac{1}{[ln(x)]^2}*\frac{d}{dx}ln(x)\]The answer is:\[f'(x)=-\frac{x}{[ln(x)]^2}\]

Answer 2

final result is wrong @D3xt3R check it

Answer 3

Ops, the correct answer is:\[f'(x)=-\frac{1}{x[ln(x)]^2}\]

Answer 4

Thanks @shahzadjalbani

Answer 5

You are welcome @D3xt3R

Answer 6

Thank you @D3xt3R

Answer 7

\[2x/\sqrt{x-1}\]
\[2x*(x-1)^-1/2\]
\[use product rule\]
\[2*(x-1)^-1/2 + -1/2(x-1)^-3/2 *2x\]
\[2/\sqrt{x-1)} -x/\sqrt{x-1}^3\]
Is my working right

Answer 8

How can i differentiate with respect to x \[2x \ln \sqrt{x+2}\]

Answer 9

apply product rule ( 2x and \(\ln\sqrt{x+2}\) )

for \(\ln\sqrt{x+2}\), you apply chain rule

Answer 10

\[(2x)' *(\ln \sqrt{x+2})+(2x)*(\ln \sqrt{x+2})'\]
\[2*(\ln \sqrt{x+2})+2x(1/x*1/2\sqrt{x+2})\]

Answer 11

actually: \((\ln\sqrt{x+2})'=(\ln\left[ (x+2)^{1/2}\right])'=\dfrac{1}{\sqrt{x+2}}\cdot\dfrac{1}{2\sqrt{x+2}}\)

Answer 12

so you just have \((\ln\sqrt{x+2})' = \dfrac{1}{2(x+2)}\)

Answer 13

does that make sense?

Answer 14

because chain rule states that \(\dfrac{d}{dx}f(~g(x)~) = f'(~g(x)~)~~g'(x)\)

Answer 15

@geerky42 so you mean its the differentiate of the whole equation - the differentiation of sqrt of x+2

Answer 16

well, I applied chain rule by let f(x) be ln x and g(x) be \(\sqrt{x+2}\).

so we have f'(g(x)) = \(\dfrac{1}{\sqrt{x+2}}\), because \((\ln x)' = \dfrac{1}{x}\), right?

then we take derivative of g(x) and have it multiplied by f'(g(x)) and we have \(g'(x) = \dfrac{1}{2}(x+2)^{-1/2}=\dfrac{1}{2\sqrt{x+2}}\)

So \(f'(g(x))\cdot g'(x) = \dfrac{1}{\sqrt{x+2}}\cdot\dfrac{1}{2\sqrt{x+2}} = \boxed{\dfrac{1}{2(x+2)}}\)

Answer 17

Thank you so much :)

Answer 18

Differentiate with respect to x
1/sin^2x

Answer 19

chain rule again: f(x) = 1/x, g(x) = sin^2 x find ( f(g(x)) )'