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Maths Differentiation Question

Can someone solve this please, I have made several attemps but can't get to the answer:

Find the second derivative of F(x)=ln(1+x^2). I can get the first derivative but cant get the second one. Show steps please!! The answer should be 2(1-x^2)/(1 + x^2)^2

Thankyou
Reply 1
First derivative:

Recall that the derivative of a logarithm, in the form ln(f(x)) ln (f(x)), is f(x)f(x)\frac{f'(x)}{f(x)}. Using this rule you get f(x)=2x1+x2f'(x) = \frac{2x}{1+x^2}. Now to get f(x)f''(x) use the Quotient rule.
Reply 2
thanyou for the response.
I can get the first derivative, the PROBLEM is the second derivative. I try and try and cant get it. If you can show it out with steps, would be great ^^
Reply 3
Original post by xiao1992
thanyou for the response.
I can get the first derivative, the PROBLEM is the second derivative. I try and try and cant get it. If you can show it out with steps, would be great ^^


To get a derivative of a function in the form f(x)h(x)\frac{f(x)}{h(x)} you basically do:

h(x)f(x)f(x)h(x)[h(x)]2\frac{h(x)f'(x)-f(x)h'(x)}{[h(x)]^2}

So for your second derivative, use f(x)=2x,h(x)=x2+1f(x) = 2x, h(x)= x^2 +1
Reply 4
Using ur formula, i get this:

2(1+x^2) - 4x^2/ (1+x^2)^2

But i dont get the answer which i posted on my first thread :s-smilie:
Original post by xiao1992
Using ur formula, i get this:

2(1+x^2) - 4x^2/ (1+x^2)^2

But i dont get the answer which i posted on my first thread :s-smilie:


That's right so far,

but now all you have to do is simplify.
So expand the numerator to get 2+2x^2 - 4x^2 so the numerator becomes 2-2x^2, then take out the common factor...
then I hope you can go on from there
:smile:

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