What is the d/(d{x)}(({f}(x,y))((2{y}+5)e^{-{x}})) ?

The d/(d{x)}(({f}(x,y))((2{y}+5)e^{-{x}})) is {f}(x,y)(2{y}+5)e^{-{x}}(-1)

What is the first d/(d{x)}(({f}(x,y))((2{y}+5)e^{-{x}})) ?

The first d/(d{x)}(({f}(x,y))((2{y}+5)e^{-{x}})) is {f}(x,y)(2{y}+5)e^{-{x}}(-1)

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Popular >

\frac{d}{d x} ((f (x, y)) ((2 y + 5) e^{- x}))

f (x, y) (2 y + 5) e^{- x} (- 1)

Solution steps

\frac{d}{d x} ((f (x, y)) ((2 y + 5) e^{- x}))

Take the constant out:

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

= f (x, y) (2 y + 5) \frac{d}{d x} (e^{- x})

Apply the chain rule:

e^{- x} \frac{d}{d x} (- x)

= e^{- x} \frac{d}{d x} (- x)

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

= f (x, y) (2 y + 5) e^{- x} (- 1)

5.77 \cdot 1 0^{- 7}

e^{- t} sin (2 t + t^{2})

[(x^{2} + 1)^{2022}]^{^{'}}

237 e^{- 1.018}

What is the d/(d{x)}(({f}(x,y))((2{y}+5)e^{-{x}})) ?
The d/(d{x)}(({f}(x,y))((2{y}+5)e^{-{x}})) is {f}(x,y)(2{y}+5)e^{-{x}}(-1)
What is the first d/(d{x)}(({f}(x,y))((2{y}+5)e^{-{x}})) ?
The first d/(d{x)}(({f}(x,y))((2{y}+5)e^{-{x}})) is {f}(x,y)(2{y}+5)e^{-{x}}(-1)