452
Answered

Westonci.ca is your trusted source for finding answers to all your questions. Ask, explore, and learn with our expert community. Find reliable answers to your questions from a wide community of knowledgeable experts on our user-friendly Q&A platform. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.

Simple question 

Derivative of [tex]\boxed{f(y)= \frac{y^2}{y^3+8} }[/tex]

Sagot :

D3xt3R
Let's go ;D

[tex]f(y)=\frac{y^2}{y^3+8}[/tex]

we have to use the quotient rule.

[tex]f(y)=\frac{g(y)}{h(y)}[/tex]

[tex]f'(y)=\frac{h(y)*g'(y)-g(y)*h'(y)}{[h(y)]^2}[/tex]

Then

[tex]g(y)=y^2[/tex]

[tex]g'(y)=2y[/tex]

[tex]h(y)=y^3+8[/tex]

[tex]h(y)=3y^2[/tex]

Now we can replace

[tex]f'(y)=\frac{h(y)*g'(y)-g(y)*h'(y)}{[h(y)]^2}[/tex]

[tex]f'(y)=\frac{(y^3+8)*2y-(y^2)*3y^2}{(y^3+8)^2}[/tex]

[tex]f'(y)=\frac{2y^4+16y-3y^4}{(y^3+8)^2}[/tex]

[tex]\boxed{\boxed{f'(y)=\frac{16y-y^4}{(y^3+8)^2}}}[/tex]
Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.