Find the derivative of the following via implicit differentiation:
d/dx(f(x)) = d/dx(1/(1+x^2))
The derivative of f(x) is f'(x):
f'(x) = d/dx(1/(1+x^2))
Using the chain rule, d/dx(1/(x^2+1)) = d/( du)1/u 0, where u = x^2+1 and ( d)/( du)(1/u) = -1/u^2:
f'(x) = -(d/dx(1+x^2))/((x^2+1)^2)
Differentiate the sum term by term:
f'(x) = -1/(1+x^2)^2 d/dx(1)+d/dx(x^2)
The derivative of 1 is zero:
f'(x) = -(d/dx(x^2)+0)/(1+x^2)^2
Simplify the expression:
f'(x) = -(d/dx(x^2))/(1+x^2)^2
Use the power rule, d/dx(x^n) = n x^(n-1), where n = 2: d/dx(x^2) = 2 x:
f'(x) = -1/(1+x^2)^2 2 x
Expand the left hand side:
Answer: |f'(x) = -(2 x)/(1+x^2)^2
+78 Firstly - convert the fraction into a linear equation;
\(1/(x^2+1)\) goes to \(1\times{(x^2+1)}^{-1}\)
Now you can just use the chain rule - or whatever you have been taught
let \(u=x^2+1\)
therefore \(f(x) = 1+ {(u)}^{-1}\)
therefore\(\frac{dy}{dx}(1+{(u)}^{-1})=-1{(u)}^{-2}\)
thats part one \o/
Part two;
if \(u=x^2+1\)
therefore \(\frac{dy}{dx}(x^2+1)=2x^1+0\)
Combine part two and part one;
\((2x)(-1){(u)}^{-2}\)
sub in for \(u\)
\((2x)(-1){(x^2+1)}^{-2}\)
simplifiy brackets
\(-2x{(x^2+1)}^{-2}\)
turn into a fraction
\(\frac{-2x}{(x^2+1){}^{2}}\)
ta dah!
