Corecto!

[Nathan23]

Use the product rule to find the derivative.

f(x) = (4x - 3)(

+ 5)

I got f'(x) = 30x + 32x^1/2 but I have a feeling I messed up my sign somewhere, is that true?

HAPPY VALENTINES DAY!

 

[stapel]

I got f'(x) = 30x + 32x^1/2

I'm sorry, but no.

You and your friends have been posting many similar exercises with similar formatting and similar tenor and within the same time-frame with no work shown. It would be very helpful (and would likely serve to legitimize your posts) if you showed what you have done.

Please reply showing your steps.

Thank you.

Eliz.

Thread edited for content.

 

[luvmykids911]

Is this my son Nathan? I think it is. I came on here to find help for those problems you gave me. You shouldn't be on the computer at school, you should be doing your work!

 

[pka]

\(\displaystyle \L
\begin{array}{l}
y = (4x - 3)(\sqrt x + 5) \\
y' = (4)(\sqrt x + 5) + (4x - 3)\left( {\frac{1}{{2\sqrt x }}} \right) \\
\end{array}\)
Now you do the work required to finish.

 

[soroban]

Hello, Nathan23!

Use the product rule to find the derivative: \(\displaystyle \:f(x)\:=\4x\,-\,3)(\sqrt{x}\,+\,5)\)

I got f'(x) = 30x + 32x^1/2 but I have a feeling I messed up my sign somewhere, is that true?
You messed up more than a sign . . .

We have: \(\displaystyle \:f(x)\;=\;(4x\,-\,3)(x^{\frac{1}{2}}\,+\,5)\)

Product Rule: \(\displaystyle \:f'(x)\:=\4x\,-\,3)\left(\frac{1}{2}x^{-\frac{1}{2}}\right) \,+ \,(x^{\frac{1}{2}}\,+\,5)\cdot4\)

Simplfy:\(\displaystyle \:f'(x)\;= \;2x^{\frac{1}{2}}\,-\,\frac{3}{2}x^{-\frac{1}{2}}\,+\,4x^{\frac{1}{2}} \,+\,20\;=\;6x^{\frac{1}{2}}\,-\,\frac{3}{2}x^{-\frac{1}{2}}\,+\,20\)