How do you differentiate $y=43^(sqrt(x))$ ?

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Answer 1 · 2016-12-14T23:19:06

$dy/dx = ( 43^sqrt(x)(ln43))/(2sqrt(x))$

Take the natural logarithm of both sides.

$lny = ln(43^sqrt(x))$

$lny = sqrt(x)(ln43)$

Differentiate:

$1/y(dy/dx) = 1/(2x^(1/2)) xx ln43 + sqrt(x)(0)$

$dy/dx = ((ln43)/(2x^(1/2)))/(1/y)$

$dy/dx = ( 43^sqrt(x)(ln43))/(2sqrt(x))$

Hopefully this helps!

How do you differentiate y=43^(sqrt(x))y=43^(sqrt(x))?