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August 28, 2014

August 28, 2014

Posted by **Jean** on Wednesday, July 18, 2012 at 6:06pm.

- Calc -
**Jean**, Wednesday, July 18, 2012 at 6:07pmCheck using the following formula:

(5^g(x))'= (ln5)*(5^g(x))*(g'(x))

- Calc -
**bobpursley**, Wednesday, July 18, 2012 at 6:22pmI 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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