Posted by Jean on Wednesday, July 18, 2012 at 6:06pm.
Check using the following formula:
(5^g(x))'= (ln5)*(5^g(x))*(g'(x))
I never did like formulas.
remember that derivative of ln(g(x))=g'(x)/g(x)
here f(x)=5^(x^2+1)
ln f(x)= (x^2+1)ln5
take the derivative
f'(x)/f(x)=2x*ln5 + 0
f'(x)=f(x)*10x*ln5=10x*ln5*5^(x^2+1)
Now comparing this to your 'formula':
f'(x)=ln5*5^(x^2+1)*2x=10x*5^(x^2+1)
Hmmm same thing.
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