Mathematics HSC 2002 10(b)(i) (1 Viewer)

[Chand]

Mathematics HSC 2002 10(b)(i)

I got stuck on this question and I've looked at the solutions that was on the main site...I got how the fraction was changed but I still don't understand how dI/dx was found. The solution is in in the attachment. Could someone please explain how it was found, and if possible, include any extra working?

Thanks,
Chand

 

[redslert]

tricky

I = (b^2 + (x + 8)^2)^-1 + (b^2 + (x - 8)^2)^-1
I = (b^2 + x^2 + 16x + 64)^-1 + (b^2 + x^2 - 16x + 64)^-1

now note that:
f(x) = f(x)^2
f'(x) = 2f(x)^1*f'(x)

.'. if I = (b^2 + x^2 + 16x + 64)^-1 + (b^2 + x^2 - 16x + 64)^-1

dI/dx = (-1)(b^2 + x^2 + 16x + 64)^-2*(2x + 16) + (-1)(b^2 + x^2 + 16x + 64)^-2*(2x - 16)

dI/dx = -2(x + 8)(b^2 + x^2 + 16x + 64)^-2 - 2(x - 8)(b^2 + x^2 + 16x + 64)^-2

dI/dx = -2(x + 8)(b^2 + (x + 8)^2)^-2 - 2(x - 8)(b^2 + (x - 8)^2)^-2

 

[Chand]

After writing it all down, I understood. Thanks for your help.