Find the first partial derivatives of the following functions. f(x,y)=(3xy+4y^{2}+1)^{5}

Tabansi

Tabansi

Answered question

2021-05-28

Find the first partial derivatives of the following functions.
f(x,y)=(3xy+4y2+1)5

Answer & Explanation

joshyoung05M

joshyoung05M

Skilled2021-05-29Added 97 answers

Step 1:Given
f(x,y)=(3xy+4y2+1)5
Step 2:Solution
f(x,y)=(3xy+4y2+1)5
fx(x,y)=5(3xy+4y2+1)51(3y)=15y(3xy+4y2+1)4fy(x,y)
=5(3xy+4y2+1)51(3x+4(2)y21)
=5(3x+8y)(3xy+4y2+1)4
Answer:
fx(x,y)=15y(3xy+4y2+1)4
fy(x,y)=5(3x+8y)(3xy+4y2+1)4
Jeffrey Jordon

Jeffrey Jordon

Expert2021-11-20Added 2605 answers

Answer is given below (on video)

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?