Question

f(x)=ln Sqaure root of 2x^2+5, find dy/dx. i am having trouble with finding out the derivative, i know that you have to make it 1/sqaure root 2x^2+5 and i know that u=(2x^2+5)^(1/2) and du=1/2(4x)^(-1/2), but after this i am confused, please help

Answer 1

If \(u = \sqrt{2x^2+5}\) then \(\displaystyle \frac{du}{dx} = \frac{4x}{2\sqrt{2x^2+5}}\) -- maybe that's where your confusion comes from.

Answer 2

On the other hand, you don't need to compute that derivative!

Answer 3

but then where does the 4x come from?

Answer 4

4x is the derivative of 2x^2 + 5.

Answer 5

but you said you dont do the derivative?

Answer 6

Right, because we can rewrite the expression before taking the derivative:
\(f(x) = \ln\sqrt{2x^2+5} = \frac12\ln(2x^2+5)\).

Answer 7

oh i see, you just take the derivative of what is in the square root, not the square root it self, and then that \[2x^{2}+5\] would be 4x and then you would times it with \[\frac{ 1 }{ \sqrt{2x ^{2}} +5}\]

Answer 8

that is supposed to be over the 5

Answer 9

Roughly, yes. You missed a 2 in the denominator but you seem to get it.

Answer 10

ok thank you

Answer 11

No problem.