What is (\partial)/(\partial x)(e^{4xy}ln(7y)) ?

The answer to (\partial)/(\partial x)(e^{4xy}ln(7y)) is ln(7y)e^{4xy}*4y

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(\partial)/(\partial x)(e^{4xy}ln(7y))

\frac{\partial}{\partial x} (e^{4 x y} ln (7 y))

ln (7 y) e^{4 x y} \cdot 4 y

Solution steps

\frac{\partial}{\partial x} (e^{4 x y} ln (7 y))

Treat

y

as a constant

Take the constant out:

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

= ln (7 y) \frac{\partial}{\partial x} (e^{4 x y})

Apply the chain rule:

e^{4 x y} \frac{\partial}{\partial x} (4 x y)

= e^{4 x y} \frac{\partial}{\partial x} (4 x y)

\frac{\partial}{\partial x} (4 x y) = 4 y

= ln (7 y) e^{4 x y} \cdot 4 y

\int_{0}^{100} 37 cos (\frac{x}{80}) d x

\int sin^{3} (4 x co) s^{2} 4 x d x

2 x y^{^{'}} + y = 8 x^{3}

(x, y) \to (0, 1) lim (x^{3} y)

What is (\partial)/(\partial x)(e^{4xy}ln(7y)) ?
The answer to (\partial)/(\partial x)(e^{4xy}ln(7y)) is ln(7y)e^{4xy}*4y