Question

Use the Product Rule to differentiate the function. Simplify your answer.
\[g(v) = (v-\sqrt{v})(v^2+\sqrt{v})\]

Answer 1

\[f(v) = (v-\sqrt{v}), h(v) = (v^2+\sqrt{v})\]
What are f'(v) and h'(v)?

Answer 2

f'(x)=1-1/2v^-1/2
h'(x) = 2v+1/2v^-1/2
???

Answer 3

Next,
\[g'(v) = f'(v)h(v) + h'(v)f(v) = (1-\frac{1}{2\sqrt{v}})(2v+\frac{1}{2\sqrt{v}})\]
Expand and simplify it?!

Answer 4

i think this is the answer:
\[3v^2-\frac{ 3 }{ 2 }\sqrt{v}-\frac{ 5 }{ 2 }v ^{3/2}-1\]

Answer 5

How do you get it? D:

Answer 6

Oh! My apology!! I did the wrong substitution!!

Answer 7

\[g'(v) = f'(v)h(v) + h'(v)f(v) = (1-\frac{1}{2\sqrt{v}}) (v^2 + \sqrt{v}) + (v-\sqrt{v})(2v+\frac{1}{2\sqrt{v}})\]
This is the right one. I'm sorry!!!

Answer 8

it's okay, i was reading my textbook and saw the formula for Product Rule

Answer 9

I just have problems dealing with the fraction exponents... :(

Answer 10

Still have problems?!

Answer 11

...

Answer 12

What do those dots mean?

Answer 13

I still got
\[g'(x)= 3v^2-\frac{ 5 }{ 2 }v ^{3/2}+\frac{ 3 }{ 2 }\sqrt{v}-1\]

Answer 14

Which is correct?!

Answer 15

Is it correct? The question mark confused me.

Answer 16

That's what I get.

Answer 17

Is there an easier way to deal with fraction exponents? sometimes i make mistakes because of exponents

Answer 18

by the way, thanks for helping me out with my derivative questions! Thank you so much! :)

Answer 19

Sometimes, factoring may help, it depends. I guess just practise more.
You're welcome :)