Question

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

Answer 1

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

Explanation:

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!