Question

differentiate
please

Answer 1

\[y=9e^x \sqrt{x}\]

Answer 2

So first, I would split it into a product of two functions:
f(x) = 9
and
\[g(x)=e^x\sqrt{x}\]

So by the product rule, the derivative would be:
\[\frac{dy}{dx} = f(x)g'(x)+g(x)f'(x)\]
But the second term ends up being 0, since the derivative of f(x) is 0 because it's a constant function.

So really we only need to worry about the first term: f(x)*g'(x)

f(x) is obviously 9, so we have:

\[9(g'(x))\]

Now remember:
\[g(x)=e^x\sqrt{x}\]
Now, g(x) can also be written as the product of two functions, we'll call them a(x) and b(x).

\[a(x)=e^x\]
\[b(x)=\sqrt{x}\]

Let's use the produce rule on g(x):

\[g'(x) = a(x)b'(x)+b(x)a'(x)\]
So:
\[g'(x) = e^x\frac{1}{2\sqrt{x}}+\sqrt{x}{e^x}\]
Remember, b'(x) is still e^x, since the derivative of e^x is e^x.

Now, the finally derivative overall is 9 * g'(x), so we have:
\[\frac{dy}{dx} = 9\left(e^x\frac{1}{2\sqrt{x}}+\sqrt{x}{e^x}\right)\]