Best solutions to - "Find both first partial derivatives. z=eysin⁡8xy"

Tazmin Horton

Answered question

2021-04-13

Find both first partial derivatives.

z = e^{y} \sin 8 x y

Answer & Explanation

Theodore Schwartz

Skilled2021-04-15Added 99 answers

Given,
$z = e^{y} \sin 8 x y$
Differentiating with respect to x, keeping y as constant, we get
$\frac{\partial z}{\partial x} = e^{y} \cdot \frac{\partial (\sin (8 x y))}{\partial x}$
$\Rightarrow \frac{\partial z}{\partial x} = e^{y} \cdot \cos (8 x y) \cdot \frac{\partial (8 x y)}{\partial x}$
$\Rightarrow \frac{\partial z}{\partial x} = e^{y} \cdot \cos (8 x y) \cdot 8 y$
$\Rightarrow \frac{\partial z}{\partial x} = 8 y e^{y} \cos (8 x y)$
Now differentiating with respect to y, keeping x as constant, we get
$\frac{\partial z}{\partial y} = \frac{\partial (e^{y})}{\partial y} \cdot \sin (8 x y) + \frac{\partial (\sin (8 x y))}{\partial y} \cdot e^{y}$
$\Rightarrow \frac{\partial z}{\partial y} = e^{y} \cdot \sin (8 x y) + \cos (8 x y) \cdot \frac{\partial (8 x y)}{\partial y} \cdot e^{y}$
$\Rightarrow \frac{\partial z}{\partial y} = e^{y} \cdot \sin (8 x y) + \cos (8 x y) \cdot 8 x \cdot e^{y}$
$\Rightarrow \frac{\partial z}{\partial y} = e^{y} [\sin (8 x y) + 8 x \cos (8 x y)]$

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Differential Calculus

Didn't find what you were looking for?