What is (\partial)/(\partial x)(e^{-(x-vt)^2}) ?

The answer to (\partial)/(\partial x)(e^{-(x-vt)^2}) is e^{-x^2+2xvt-v^2t^2}(-2x+2vt)

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(\partial)/(\partial x)(e^{-(x-vt)^2})

\frac{\partial}{\partial x} (e^{- (x - v t)^{2}})

e^{- x^{2} + 2 xv t - v^{2} t^{2}} (- 2 x + 2 v t)

Solution steps

\frac{\partial}{\partial x} (e^{- (x - v t)^{2}})

Simplify

e^{- (x - v t)^{2}} : e^{- x^{2} + 2 xv t - v^{2} t^{2}}

= \frac{\partial}{\partial x} (e^{- x^{2} + 2 xv t - v^{2} t^{2}})

Treat

v, t

as constants

Apply the chain rule:

e^{- x^{2} + 2 xv t - v^{2} t^{2}} \frac{\partial}{\partial x} (- x^{2} + 2 xv t - v^{2} t^{2})

= e^{- x^{2} + 2 xv t - v^{2} t^{2}} \frac{\partial}{\partial x} (- x^{2} + 2 xv t - v^{2} t^{2})

\frac{\partial}{\partial x} (- x^{2} + 2 xv t - v^{2} t^{2}) = - 2 x + 2 v t

= e^{- x^{2} + 2 xv t - v^{2} t^{2}} (- 2 x + 2 v t)

\frac{\partial}{\partial x} (x cos (x) sin (x))

f (x) = cos (x) sec (x)

x^{2} - 5

\frac{d}{d x} ((2 x)^{3 x})

\frac{d}{d x} (5 x - 9)

What is (\partial)/(\partial x)(e^{-(x-vt)^2}) ?
The answer to (\partial)/(\partial x)(e^{-(x-vt)^2}) is e^{-x^2+2xvt-v^2t^2}(-2x+2vt)