452
Answered

Westonci.ca is the Q&A platform that connects you with experts who provide accurate and detailed answers. Explore in-depth answers to your questions from a knowledgeable community of experts across different fields. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.

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 trusting us with your questions. We're here to help you find accurate answers quickly and efficiently. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.