What is (\partial)/(\partial x)(2x+3y+6xye^{3x^2y}) ?

The answer to (\partial)/(\partial x)(2x+3y+6xye^{3x^2y}) is 6y(e^{3x^2y}+6e^{3x^2y}x^2y)+2

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(\partial)/(\partial x)(2x+3y+6xye^{3x^2y})

\frac{\partial}{\partial x} (2 x + 3 y + 6 x y e^{3 x^{2} y})

6 y (e^{3 x^{2} y} + 6 e^{3 x^{2} y} x^{2} y) + 2

Solution steps

\frac{\partial}{\partial x} (2 x + 3 y + 6 x y e^{3 x^{2} y})

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (2 x) + \frac{\partial}{\partial x} (3 y) + \frac{\partial}{\partial x} (6 x y e^{3 x^{2} y})

\frac{\partial}{\partial x} (2 x) = 2

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

\frac{\partial}{\partial x} (6 x y e^{3 x^{2} y}) = 6 y (e^{3 y x^{2}} + 6 y e^{3 y x^{2}} x^{2})

= 2 + 0 + 6 y (e^{3 y x^{2}} + 6 y e^{3 y x^{2}} x^{2})

Simplify

= 6 y (e^{3 x^{2} y} + 6 e^{3 x^{2} y} x^{2} y) + 2

x \to - 1 lim (- 6)

d er i v a t i v e 9 x - (sin^{2} (2 x))

f (x) = arccsc (x)

t an g e n t f (x) = ln (tan (2 x)), at x = \frac{π}{8}

n = 1 \sum \infty \frac{( - 3 ) ^{n}}{n}

What is (\partial)/(\partial x)(2x+3y+6xye^{3x^2y}) ?
The answer to (\partial)/(\partial x)(2x+3y+6xye^{3x^2y}) is 6y(e^{3x^2y}+6e^{3x^2y}x^2y)+2