How do you differentiate #y=(4x^5+x^2+4)/(5x^2-2)# using the quotient rule?

Question

Samantha Adkison · Accepted Answer

#(60x^6-40x^4-44x)/(5x^2-2)^2#

When we want to derive the following y=#(4x^5+x^2+4)/(5x^2-2)# using the quotient rule we follow the format of #((f')(g)-(g')(f))/(g^2)#. The top is our #f# and the bottom is #g#, since this is just a polynomial we use the power rule #nx^(n-1)#. We take the #d/dx# of the top and multiply it by the bottom then we subtract the #d/dx# of the bottom mulitplied by the top and we square the bottom. Just make sure to derive the polynomial right then multiply and combine like terms.

This is what you should end up when you derive the equation.

#(20x^4+2x(5x^2-2)-10x(4x^5+x^2+4))/((5x^2-2)^2)#

Then distribute the top and combine like terms.

#(100x^6-40x^4+10x^3-4x-40x^6-10x^3-40x)/(5x^2-2)^2#

Note: #10x^3-10x^3# cancel out. We leave the bottom how it is don't try to FOIL it out.

#(60x^6-40x^4-44x)/(5x^2-2)^2#

This is our final answer :) If you have any questions don't be afraid to ask your boii.

Isabella Ackerman · Answer

undefined To differentiate $ y = \frac{4x^5 + x^2 + 4}{5x^2 - 2} $ using the quotient rule, the formula is:

$$ \frac{d}{dx} \left( \frac{f(x)}{g(x)} \right) = \frac{g(x) \cdot f'(x) - f(x) \cdot g'(x)}{(g(x))^2} $$

Where $ f(x) = 4x^5 + x^2 + 4 $ and $ g(x) = 5x^2 - 2 $.

Now, differentiate $ f(x) $ and $ g(x) $ to get $ f'(x) $ and $ g'(x) $, respectively. Then, substitute into the quotient rule formula and simplify.

$$ f'(x) = 20x^4 + 2x $$

$$ g'(x) = 10x $$

Now, substitute these into the quotient rule formula:

$$ \frac{(5x^2 - 2)(20x^4 + 2x) - (4x^5 + x^2 + 4)(10x)}{(5x^2 - 2)^2} $$

Simplify this expression to get the derivative of $ y $ with respect to $ x $.