How do you differentiate $y = 3^{2 x + 7}$ $y= 3^(2x+7)$ ?

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Answer 1 · 2015-03-30T02:35:38

Use $\frac{d}{d x} (a^{x}) = a^{x} ln a$ $d/(dx)(a^x)=a^x ln a$ together with the chain rule:

$d/(dx)(f(u))=f'(u) (du)/(dx)$

For $y=3^(2x+7)$ we get:

$(dy)/(dx) = 3^(2x+7) ln3 (2) =2(3^(2x+7) ) ln3$

Or use $2ln3=ln9$ to write:

$(dy)/(dx) = 3^(2x+7) ln9$

How do you differentiate y= 3^(2x+7)y=32x+7y= 3^(2x+7)?