What is (\partial)/(\partial x)(-3cos(pix)+6x^2y+5e^{xz}+3yz-98) ?

The answer to (\partial)/(\partial x)(-3cos(pix)+6x^2y+5e^{xz}+3yz-98) is 12xy+5e^{xz}z+3pisin(pix)

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(\partial)/(\partial x)(-3cos(pix)+6x^2y+5e^{xz}+3yz-98)

\frac{\partial}{\partial x} (- 3 cos (π x) + 6 x^{2} y + 5 e^{x z} + 3 yz - 98)

12 x y + 5 e^{x z} z + 3 π sin (π x)

Solution steps

\frac{\partial}{\partial x} (- 3 cos (π x) + 6 x^{2} y + 5 e^{x z} + 3 yz - 98)

Treat

y, z

as constants

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= - \frac{\partial}{\partial x} (3 cos (π x)) + \frac{\partial}{\partial x} (6 x^{2} y) + \frac{\partial}{\partial x} (5 e^{x z}) + \frac{\partial}{\partial x} (3 yz) - \frac{\partial}{\partial x} (98)

\frac{\partial}{\partial x} (3 cos (π x)) = - 3 π sin (π x)

\frac{\partial}{\partial x} (6 x^{2} y) = 12 y x

\frac{\partial}{\partial x} (5 e^{x z}) = 5 e^{x z} z

\frac{\partial}{\partial x} (3 yz) = 0

\frac{\partial}{\partial x} (98) = 0

= - (- 3 π sin (π x)) + 12 y x + 5 e^{x z} z + 0 - 0

Simplify

= 12 x y + 5 e^{x z} z + 3 π sin (π x)

\int x - \frac{8}{x} d x

ln^{2} (1 + x)

n = 1 \sum \infty \frac{9}{π ^{n}}

\int \frac{1}{361 - x ^{2}} d x

\frac{d y}{d x} - \frac{y}{x} = x e^{x}, y (1) = e - 1

What is (\partial)/(\partial x)(-3cos(pix)+6x^2y+5e^{xz}+3yz-98) ?
The answer to (\partial)/(\partial x)(-3cos(pix)+6x^2y+5e^{xz}+3yz-98) is 12xy+5e^{xz}z+3pisin(pix)