Quotient rule

[Bobby Jones]

Using the quotient rule, find the gradient of the function:

y = (2x) / (5x^2 - 2x + 1)^1/2


Am I correct with this answer:

dy/dx = ( 2(5x^2-2x+1)^1/2 - x/((10x - 2)^1/2)) / (5x^2 - 2x + 1)


What software do you guys use to write the formulas properly?

 

[soroban]

Hello, Bobby Jones!

Using the quotient rule, find the gradient of the function:

. . \(\displaystyle y \:=\: \frac{2x}{(5x^2 - 2x + 1)^{\frac{1}{2}}}\)

Am I correct with this answer:

\(\displaystyle \frac{dy}{dx} \;=\; \frac{2(5x^2-2x+1)^{\frac{1}{2}} - \frac{x}{(10x - 2)^{\frac{1}{2}}}} {5x^2 - 2x + 1}\) . Sorry, no . . .

\(\displaystyle \text{We have: }\:y \;=\;\frac{\overbrace{2x}^f}{\underbrace{(5x^2-2x+1)^{\frac{1}{2}}}_g}\)

\(\displaystyle \text{Then: }\:\frac{dy}{dx} \;=\;\frac{\overbrace{(5x^2-2x+1)^{\frac{1}{2}}}^g \cdot \overbrace{2}^{f'} \;-\; \overbrace{2x}^f\cdot\overbrace{\tfrac{1}{2}(5x^2-2x+1)^{\text{-}\frac{1}{2}}(10x-2)}^{g'}}{\underbrace{5x^2-2x+1}_{g^2}}\)

. . . . . . .\(\displaystyle =\;\frac{2(5x^2-2x+1)^{\frac{1}{2}} - 2x(5x-1)(5x^2 - 2x + 1)^{\text{-}\frac{1}{2}}}{5x^2-2x+1}\)

\(\displaystyle \text{Factor: }\;=\;\frac{2(5x^2-2x+1)^{\text{-}\frac{1}{2}}\cdot\bigg[(5x^2-2x+1) - x(5x-1)\bigg]}{5x^2-2x + 1}\)

. . . . . . .\(\displaystyle =\;\frac{2\left(5x^2 - 2x + 1 - 5x^2 + x\right)}{(5x^2-2x+1)^{\frac{3}{2}}}\)

. . . . . . .\(\displaystyle =\;\frac{2(1-x)}{(5x^2 - 2x + 1)^{\frac{3}{2}}}\)

 

[galactus]

What software do you guys use to write the formulas properly?

It isn't software. You can do it too. Type the LaTex code. To see how Soroban displayed his nice format, click on 'quote' in the upper right corner of his post.

He does a meticulous job. Most do not take the time to be so eloquent.

For instance, to display \(\displaystyle \int_{0}^{1}e^{2x}dx\), type \int_{0}^{1}e^{-2x}dx surrounded with the tag \(\displaystyle . The tag when closing must have a forward slash, /tex, in brackets.\)

 

[Bobby Jones]

Cheers Soroban, I can see where I made my mistake now, nicely displayed aswell. And Thanks galactus, I'll try and use the LaTex code stuff in future.