What is (\partial)/(\partial x)(e^{-(x^3+y^3)}) ?

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

(\partial)/(\partial x)(e^{-(x^3+y^3)})

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

- 3 e^{- x^{3} - y^{3}} x^{2}

Solution steps

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

Simplify

e^{- (x^{3} + y^{3})} : e^{- x^{3} - y^{3}}

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

Treat

y

as a constant

Apply the chain rule:

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

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

\frac{\partial}{\partial x} (- x^{3} - y^{3}) = - 3 x^{2}

= e^{- x^{3} - y^{3}} (- 3 x^{2})

Simplify

= - 3 e^{- x^{3} - y^{3}} x^{2}

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

t an g e n t f (x) = x^{3} - 5 x + 3, at x = 3

d er i v a t i v e f (x) = (- x^{2} - 4)^{2}

\frac{d}{d x} (3 x^{2} - \frac{1}{24} ln (x))

\int \frac{1}{x ^{2} + 4 x + 20} d x

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