# Math Help - Help with derivatives

1. ## Help with derivatives

Hi, I'm not sure if I'm doing the right way. Here is the problems

1. sqrt (6x+5) (Get the derivative using this formulate f(x) = ((x+h) - x)/h ) lim h-> 0 so this is my attemp.

[ sqrt(6(x+h)+5) - sqrt(6x+5) ] / h

then I rationalize the numerator

(6x+6h+5 - 6x-5 ) / h(sqrt(6x+6h+5)(sqrt(6x+5)

then I simplify the numberator

6h / h(sqrt(6x+6h+5)(sqrt(6x+5)

can you tell me what i am doing the right way or show me how to do it.

2. the second is the product rule of derivative (sqrt(x) + 3x)(5x^2 - 3/x)

apply the rule,

(1/2 x^-1/2 + 3)(5x^2 - 3/x) + (sqrt(x) + 3x)(10x + 3/x^2)

then i put down the negative power to denominator
(1/(2 sqrt(x)) + 3)(5x^2 - 3/x) + (sqrt(x) + 3x)(10x + 3/x^2)

then what should i do next? or show me the next step.

thanks

2. ## Re: Help with derivatives

For 1 you have done everything correctly, so now cancel off the factor of h from the top and bottom and then make h -> 0.

For 2 you have done enough. If you REALLY want to you could get a common denominator and try to simplify, but the idea is that you are trying to find a correct derivative, which you have already done.

3. ## Re: Help with derivatives

Originally Posted by Prove It
For 1 you have done everything correctly, so now cancel off the factor of h from the top and bottom and then make h -> 0.

For 2 you have done enough. If you REALLY want to you could get a common denominator and try to simplify, but the idea is that you are trying to find a correct derivative, which you have already done.
I manage to get the answer in 2. but what do you mean cancel off the factor of h? can you show me. thanks

4. ## Re: Help with derivatives

You've ended up with \displaystyle \begin{align*} \frac{6h}{h\left( \sqrt{6x + 6h + 5} + \sqrt{6x + 5} \right) } \end{align*}. Surely you can see something you can cancel...

5. ## Re: Help with derivatives

Originally Posted by Prove It
You've ended up with \displaystyle \begin{align*} \frac{6h}{h\left( \sqrt{6x + 6h + 5} + \sqrt{6x + 5} \right) } \end{align*}. Surely you can see something you can cancel...
can i cancel the h in top and bottom, then limit h->0 then i will have 6 / ((sqrt (6x + 5) + sqrt(6x+5)) . and is there a + sign in th denominator of two sqrt?

sorry. im not really good at math

6. ## Re: Help with derivatives

Yes that's right. And yes you need the + sign there, because to rationalise the numerator of \displaystyle \begin{align*} \sqrt{6x + 6h + 5} - \sqrt{6x + 5} \end{align*} you need to multiply top and bottom by \displaystyle \begin{align*} \sqrt{6x + 6h + 5} + \sqrt{ 6x + 5} \end{align*}.

7. ## Re: Help with derivatives

Originally Posted by Prove It
Yes that's right. And yes you need the + sign there, because to rationalise the numerator of \displaystyle \begin{align*} \sqrt{6x + 6h + 5} - \sqrt{6x + 5} \end{align*} you need to multiply top and bottom by \displaystyle \begin{align*} \sqrt{6x + 6h + 5} + \sqrt{ 6x + 5} \end{align*}.
thanks alot.

8. ## Re: Help with derivatives

Originally Posted by Prove It
Yes that's right. And yes you need the + sign there, because to rationalise the numerator of \displaystyle \begin{align*} \sqrt{6x + 6h + 5} - \sqrt{6x + 5} \end{align*} you need to multiply top and bottom by \displaystyle \begin{align*} \sqrt{6x + 6h + 5} + \sqrt{ 6x + 5} \end{align*}.
I thought it's okay to plug in h=0 at this point though... Although it's technically h->0, what i was taught with was, you can straight up plug in h=0 if nothing will go wrong because of that.