# Derivative ^x, is this correct?

## Homework Statement

$$y=\sqrt{x} * 19^x$$

## The Attempt at a Solution

$$y=x^{1/2}*19^x$$
$$y'=1/2x^{-1/2}*19$$

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You need to use the product rule. If you didn't know:
$$19^x = e^{\ln 19^x} = e^{x\ln 19} = e^{(\ln 19)x}$$

I'm double-checking everything now.

HallsofIvy
Homework Helper

## Homework Statement

$$y=\sqrt{x} * 19^x$$

## The Attempt at a Solution

$$y=x^{1/2}*19^x$$
$$y'=1/2x^{-1/2}*19$$
No, that's not right and I don't see how you could have got that! The derivative of a product is typically a sum of derivatives and the dervative of "$19^x$" is definitely not 19.

I would just take the logarithm of both sides immediately:
$$ln(y)= ln(\sqrt{x}19^x)= ln(x^{1/2})+ ln(19^x)$$
$$ln(y)= (1/2)ln(x)+ x ln(19)$$

Now differentiate both sides and solve for y'.

y = sqrt(x) * 10^x

y = sqrt(x) * (10^x) + 10^x * (sqrt(x)) `