1. ## Differentiate questions

Hi

I need help on the following two questions:

1)Differentiate with respect with x: $\frac{1}{(7x^2+3)^5}$

This is what i have done, not sure if i have done it correctly:

using quotient rule

$\frac{(7x^2+3)^5 * 0- 1 * 5(7x^2+3)^4 * 14x}{((7x^2+3)^5)^2}$

$\frac{-70x(7x^2+3)^4}{(7x^2+3)^6}$

$\frac{-70x}{(7x^2+3)^6}$

2)Differentiate with respect with x: $(2x-1)^4 * (x+5)^{\frac{1}{2}}$

using the product rule

$(2x-1)^4 * \frac{1}{2}(x+5)^{\frac{1}{2}} + (x+5)^{\frac{1}{2}} * 4(2x-1)^3 * 2$

$\frac{2(2x-1)^4}{\sqrt{x+5}} + \frac{8(2x-1)^3 * (x+5)^{\frac{1}{2}}}{\sqrt{x+5}}$

What would i do next?

P.S

2. Originally Posted by Paymemoney
Hi

I need help on the following two questions:

1)Differentiate with respect with x: $\frac{1}{(7x^2+3)^5}$

This is what i have done, not sure if i have done it correctly:

using quotient rule

$\frac{(7x^2+3)^5 * 0- 1 * 5(7x^2+3)^4 * 14x}{((7x^2+3)^5)^2}$

$\frac{-70x(7x^2+3)^4}{(7x^2+3)^6}$

$\frac{-70x}{(7x^2+3)^6}$

2)Differentiate with respect with x: $(2x-1)^4 * (x+5)^{\frac{1}{2}}$

using the product rule

$(2x-1)^4 * \frac{1}{2}(x+5)^{\frac{1}{2}} + (x+5)^{\frac{1}{2}} * 4(2x-1)^3 * 2$

$\frac{2(2x-1)^4}{\sqrt{x+5}} + \frac{8(2x-1)^3 * (x+5)^{\frac{1}{2}}}{\sqrt{x+5}}$

What would i do next?

P.S
Well for question 1, just bring the denominator up to the top line where it becomes $(7x^{2}+3)^{-5}$. Then just use the chain rule... However your solution using the quotient rule is correct

On question 2, your forgeting to reduce the power by 1 when differentiating
$( x + 5 )^{\frac{1}{2}}$...

Are you familar with how the chain rule works?

3. ok i redone it and this is what i got:

$\frac{(2x-1)^4}{2\sqrt{x+5}} + \frac{16(x+5)(2x-1)^3}{2\sqrt{x+5}}$

Would i expand it??

4. well when i worked it out i got

$\frac{dy}{dx} = \frac{(2x-1)^{4}}{2\sqrt{x+5}} + 8(2x - 1)^3.\sqrt{x+5}$

There is no need to expand that but i guess you could tidy it up a little by using $2\sqrt{x+5}$ as your common denominator...

5. Well how did the book's answer get this:

$\frac{(2x-1)^3(18x+79)}{2\sqrt{x+5}}$

6. Originally Posted by Paymemoney
Well how did the book's answer get this:

$\frac{(2x-1)^3(18x+79)}{2\sqrt{x+5}}$
Read the last sentence of post #4 again.

7. Originally Posted by Paymemoney
Well how did the book's answer get this:

$\frac{(2x-1)^3(18x+79)}{2\sqrt{x+5}}$
Ok i go through this with you...

Using the common denominator $2\sqrt{x+5}$

you get $\frac{dy}{dx} = \frac{(2x-1)^{4} + 2\sqrt{x+5}(8(2x-1)^{3}.\sqrt{x+5})}{2\sqrt{x+5}}$

Tidying up this you get

$\frac{dy}{dx} = \frac{(2x-1)^{4} + 16(2x-1)^{3}.(x+5)}{2\sqrt{x+5}}$ as $\sqrt{x+5}.\sqrt{x+5} = x+5$

then taking out $(2x-1)^{3}$ as a common factor of the top line

you get $\frac{dy}{dx} = \frac{(2x-1)^3(2x-1+16x+80)}{2\sqrt{x+5}}$

so tidying up what's in the brackets leaves you with

$\frac{dy}{dx} = \frac{(2x-1)^{3}(18x+79)}{2\sqrt{x+5}}$