What is (\partial)/(\partial x)(3xe^{7xy}) ?

The answer to (\partial)/(\partial x)(3xe^{7xy}) is 3(e^{7xy}+7e^{7xy}xy)

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(\partial)/(\partial x)(3xe^{7xy})

\frac{\partial}{\partial x} (3 x e^{7 x y})

3 (e^{7 x y} + 7 e^{7 x y} x y)

Solution steps

\frac{\partial}{\partial x} (3 x e^{7 x y})

Treat

y

as a constant

Take the constant out:

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

= 3 \frac{\partial}{\partial x} (x e^{7 x y})

Apply the Product Rule:

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

f = x, g = e^{7 x y}

= 3 (\frac{\partial}{\partial x} (x) e^{7 x y} + \frac{\partial}{\partial x} (e^{7 x y}) x)

\frac{\partial}{\partial x} (x) = 1

\frac{\partial}{\partial x} (e^{7 x y}) = e^{7 x y} \cdot 7 y

= 3 (1 \cdot e^{7 x y} + e^{7 x y} \cdot 7 y x)

Simplify

= 3 (e^{7 x y} + 7 e^{7 x y} x y)

\int_{0}^{7} \frac{4 t}{( t - 8 ) ^{2}} d t

n = 0 \sum \infty \frac{1}{n ^{3}}

\frac{d}{d x} (\frac{4}{x - 1})

\int (x - 1)^{2} x d x

\frac{64}{( 1 + 4 x ) ^{3}}

What is (\partial)/(\partial x)(3xe^{7xy}) ?
The answer to (\partial)/(\partial x)(3xe^{7xy}) is 3(e^{7xy}+7e^{7xy}xy)