Tiffany Russell

2022-01-14

Compute the given partial derivatives.
$f\left(x,y\right)=3{x}^{2}y+4{x}^{3}{y}^{2}-7x{y}^{5},{f}_{x}\left(1,2\right)$

eskalopit

Expert

Step 1
Given,
$f\left(x,y\right)=3{x}^{2}y+4{x}^{3}{y}^{2}-7x{y}^{5},{f}_{x}\left(1,2\right)$
The objective is to determine the partial derivatives ${f}_{x}$ of the given function at given point.
For partial differentiation, differentiate the given function with respect to x, taking y as a constant.
Step 2
${f}_{x}=\frac{\partial }{\partial x}\left(3{x}^{2}y+4{x}^{3}{y}^{2}-7x{y}^{5}\right)$
$=\frac{\partial }{\partial x}\left(3{x}^{2}y\right)+\frac{\partial }{\partial x}\left(4{x}^{3}{y}^{2}\right)-\frac{\partial }{\partial x}\left(7x{y}^{5}\right)$
$=6yx+12{y}^{2}{x}^{2}-7{y}^{5}$
Thus, the value of ${f}_{x}$ at (1,2) is,
${f}_{x}\left(1,2\right)=6yx+12{y}^{2}{x}^{2}-7{y}^{5}$
$=6\left(2\right)\left(1\right)+12{\left(2\right)}^{2}{\left(1\right)}^{2}-7{\left(2\right)}^{5}$
$=6\left(2\right)\left(1\right)+12\cdot 4{\left(1\right)}^{2}-7{\left(2\right)}^{5}$
=12+12*4*1-7*32
=12+48-224
=-164

Do you have a similar question?