Get the solution to "Find all the second partial derivatives f(x,y)=sin2(mx+ny)"

Tara Alvarado

Answered question

2021-12-23

Find all the second partial derivatives

f (x, y) = \sin^{2} (m x + n y)

Answer & Explanation

GaceCoect5v

Beginner2021-12-24Added 26 answers

f (x, y) = \sin^{2} (m x + n y)

f_{x} = 2 \sin (m x + n y) \times \frac{\partial}{\partial x} [\sin (m x + n y)]

= 2 m \sin (m x + n y) \times \cos (m x + n y)

Similary

f_{y} = 2 n \sin (m x + n y) \times \cos (m x + n y)

\Rightarrow f_{x} = m \sin (2 (m x + n y))

f_{y} = n \sin (2 (m x + n y))

f_{x x} = m \cdot \frac{\partial}{\partial x} \sin (2 (m x + n y))

= 2 m \times m \times \cos (2 (m x + n y))

f_{x x} = 2 m^{2} \cos (2 (m x + n y))

f_{y y} = n \cdot \frac{\partial}{\partial y} \sin (2 (m x + n y))

= 2 n \times n \times \cos (2 (m x + n y))

f_{y y} = 2 n^{2} \cos (2 (m x + n y))

recoronarrv

Beginner2021-12-25Added 20 answers

If you mean $f (x, y) = \sin^{2} (m x + n y) :$
$f_{x} = 2 \sin (m x + n y) \cdot m \cos (m x + n y) = m \sin (2 (m x + n y))$ , via double angle identity
$f_{y} = 2 \sin (m x + n y) \cdot n \cos (m x + n y) = n \sin (2 (m x + n y))$
$f_{x x} = m \cdot 2 m \cos (2 m x + 2 n y) = 2 m^{2} \cos (2 m x + 2 n y)$
$f_{x y} = m \cdot 2 n \cos (2 m x + 2 n y) = 2 m n \cos (2 m x + 2 n y)$
$f_{y y} = n \cdot 2 n \cos (2 m x + 2 n y) = 2 n^{2} \cos (2 m x + 2 n y)$ .

user_27qwe

Skilled2021-12-30Added 375 answers

this is the best answer and the other one is not

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Differential Calculus

Didn't find what you were looking for?