How do you differentiate $t(x)= 3^(7x-3)$ ?

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Answer 1 · 2016-11-20T15:07:10

Take the natural logarithm of each side.

$y = 3^(7x - 3)$

$lny = ln(3^(7x- 3))$

$lny = (7x - 3)ln3$

Differentiate using the product rule and the rule that $(lnx)' = 1/x$ .

$1/y(dy/dx) = 7ln3 + (7x - 3)0$

$1/y(dy/dx) = 7ln3$

$dy/dx = 7ln3/(1/y)$

$dy/dx= yln3^7$

$dy/dx= 3^(7x- 3)ln2187$

Hopefully this helps!

How do you differentiate t(x)= 3^(7x-3)t(x)= 3^(7x-3)?