What is (\partial)/(\partial x)(x^2y^2e^{6xy}) ?

The answer to (\partial)/(\partial x)(x^2y^2e^{6xy}) is y^2(2xe^{6xy}+e^{6xy}*6yx^2)

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(\partial)/(\partial x)(x^2y^2e^{6xy})

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

y^{2} (2 x e^{6 x y} + e^{6 x y} \cdot 6 y x^{2})

Solution steps

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

Treat

y

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x^{2}, g = e^{6 x y}

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

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

\frac{\partial}{\partial x} (e^{6 x y}) = e^{6 x y} \cdot 6 y

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

\int \frac{3 - 18 x}{16 - 9 x ^{2}} d x

y^{^{''}} + y^{^{'}} - 2 y = e^{- 2 t}

t \to t lim (3 t^{\frac{1}{3}})

\frac{d}{d x} ((2 x + 1)^{5} (x^{3} - x + 1)^{4})

\int 2 - 2 cos (x) d x

What is (\partial)/(\partial x)(x^2y^2e^{6xy}) ?
The answer to (\partial)/(\partial x)(x^2y^2e^{6xy}) is y^2(2xe^{6xy}+e^{6xy}*6yx^2)