How do I find the derivative of this function:
2x/[(x^2+1)^1/2]?
I'm having such I hard time finding this derivative:s I've spent over an hour trying to get the textbook answer of f'(x)=2/(x^2+1)^3/2.
Thanks(:
2x/[(x^2+1)^1/2]?
I'm having such I hard time finding this derivative:s I've spent over an hour trying to get the textbook answer of f'(x)=2/(x^2+1)^3/2.
Thanks(:
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ok let's mae it together:
f (x) = [ 2(x^2+1)^1/2 - 2x[ 2x/ 2(x^2+1)^1/2] ] / (x^2 + 1)
... to simplify well multiply by (x^2+1)^1/2 / (x^2+1)^1/2 and then u get :
f '(x) = [ 2(x^2 + 1) - 2x^2] / (x^2 + 1)^3/2 ... u can easily show that
s f '(x) = (2x^2 + 2 - 2x^2) / / (x^2 + 1)^3/2
...f '(x) = 2 / (x^2 + 1)^3/2
Here it's proved in few steps.
f (x) = [ 2(x^2+1)^1/2 - 2x[ 2x/ 2(x^2+1)^1/2] ] / (x^2 + 1)
... to simplify well multiply by (x^2+1)^1/2 / (x^2+1)^1/2 and then u get :
f '(x) = [ 2(x^2 + 1) - 2x^2] / (x^2 + 1)^3/2 ... u can easily show that
s f '(x) = (2x^2 + 2 - 2x^2) / / (x^2 + 1)^3/2
...f '(x) = 2 / (x^2 + 1)^3/2
Here it's proved in few steps.
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lodhi/hidlo