Thread: DERIVATE OF A INVERSE FUNCTION...

1. DERIVATE OF A INVERSE FUNCTION...

$y=\frac{1}{2x^2+3x-5}$ find y'

could someone help me...no idea what to do...

2. Originally Posted by juanfe_zodiac
$y=\frac{1}{2x^2+3x-5}$ find y'

could someone help me...no idea what to do...
Have you been taught the quotient rule? Use it here if you have.

Alternatively, since $y = (2x^2 + 3x - 5)^{-1}$ you can also use the chain rule. Have you been taught it?

3. Originally Posted by juanfe_zodiac
$y=\frac{1}{2x^2+3x-5}$ find y'

could someone help me...no idea what to do...
By the way, this is NOT an "inverse function", it is a reciprocal.

4. Originally Posted by mr fantastic
Have you been taught the quotient rule? Use it here if you have.

Alternatively, since $y = (2x^2 + 3x - 5)^{-1}$ you can also use the chain rule. Have you been taught it?

from $y = (2x^2 + 3x - 5)^{-1}$ i got

$(-x^{-2})(2x+3)(1)$ where $x=(2x^2 + 3x - 5)$ for the first parenthesis.

is it correct? i dont know hoe to continue it..help plz..i tried but i couldnt get it..tks

5. Originally Posted by juanfe_zodiac
from $y = (2x^2 + 3x - 5)^{-1}$ i got

$(-x^{-2})(2x+3)(1)$ where $x=(2x^2 + 3x - 5)$ for the first parenthesis.

is it correct? Unfortunately No!
i dont know hoe to continue it..help plz..i tried but i couldnt get it..tks
$f(x)=\dfrac1{2x^2+3x-5}=\left(2x^2+3x-5\right)^{-1}$

Let $u(x) = 2x^2+3x-5$ then $f(x)=(u(x))^{-1}$

You have to use the chain rule to get the derivation:

$f'(x) = (-1)\cdot (u(x))^{-2}\cdot u'(x)$

$f(x)=(-1)\cdot (2x^2+3x-5)^{-2} \cdot (4x+3)$
$f(x)=-\dfrac{4x+3}{(2x^2+3x-5)^2}$