452
Answered

Discover a wealth of knowledge at Westonci.ca, where experts provide answers to your most pressing questions. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.

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]
Thank you for choosing our service. We're dedicated to providing the best answers for all your questions. Visit us again. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.