What is (\partial)/(\partial x)(5e^{x^2-4y}) ?

The answer to (\partial)/(\partial x)(5e^{x^2-4y}) is 10e^{x^2-4y}x

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(\partial)/(\partial x)(5e^{x^2-4y})

\frac{\partial}{\partial x} (5 e^{x^{2} - 4 y})

10 e^{x^{2} - 4 y} x

Solution steps

\frac{\partial}{\partial x} (5 e^{x^{2} - 4 y})

Treat

y

as a constant

Take the constant out:

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

= 5 \frac{\partial}{\partial x} (e^{x^{2} - 4 y})

Apply the chain rule:

e^{x^{2} - 4 y} \frac{\partial}{\partial x} (x^{2} - 4 y)

= e^{x^{2} - 4 y} \frac{\partial}{\partial x} (x^{2} - 4 y)

\frac{\partial}{\partial x} (x^{2} - 4 y) = 2 x

= 5 e^{x^{2} - 4 y} \cdot 2 x

Simplify

= 10 e^{x^{2} - 4 y} x

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

n = 1 \sum \infty n! e^{- n}

\frac{d y}{d x} = y^{2} sin (x^{2})

x \to - 1 - lim (\frac{3}{x ^{2} - 1})

\frac{\partial}{\partial x} (\frac{5 x y}{x - y})

What is (\partial)/(\partial x)(5e^{x^2-4y}) ?
The answer to (\partial)/(\partial x)(5e^{x^2-4y}) is 10e^{x^2-4y}x