What is the derivative of f(x)=sqrt(x)(5e^x-4x^4) ?

The derivative of f(x)=sqrt(x)(5e^x-4x^4) is (-36x^4+10e^xx+5e^x)/(2sqrt(x))

What is the first derivative of f(x)=sqrt(x)(5e^x-4x^4) ?

The first derivative of f(x)=sqrt(x)(5e^x-4x^4) is (-36x^4+10e^xx+5e^x)/(2sqrt(x))

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derivative of

f (x) = x (5 e^{x} - 4 x^{4})

\frac{- 36 x ^{4} + 10 e ^{x} x + 5 e ^{x}}{2 x}

Solution steps

\frac{d}{d x} (x (5 e^{x} - 4 x^{4}))

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x, g = 5 e^{x} - 4 x^{4}

= \frac{d}{d x} (x) (5 e^{x} - 4 x^{4}) + \frac{d}{d x} (5 e^{x} - 4 x^{4}) x

\frac{d}{d x} (x) = \frac{1}{2 x}

\frac{d}{d x} (5 e^{x} - 4 x^{4}) = 5 e^{x} - 16 x^{3}

= \frac{1}{2 x} (5 e^{x} - 4 x^{4}) + (5 e^{x} - 16 x^{3}) x

Simplify

\frac{1}{2 x} (5 e^{x} - 4 x^{4}) + (5 e^{x} - 16 x^{3}) x : \frac{- 36 x ^{4} + 10 e ^{x} x + 5 e ^{x}}{2 x}

= \frac{- 36 x ^{4} + 10 e ^{x} x + 5 e ^{x}}{2 x}

\int 2^{x^{2}} d x

\frac{1}{( s ^{2} + 6 π s + π ^{2} )}

((x - 4)^{2} + \frac{1}{2}), [2, 4]

\int \frac{4 x + 10}{x ^{3} + x} d x

\frac{d y}{d x} + 0.7 y = 18.2

What is the derivative of f(x)=sqrt(x)(5e^x-4x^4) ?
The derivative of f(x)=sqrt(x)(5e^x-4x^4) is (-36x^4+10e^xx+5e^x)/(2sqrt(x))
What is the first derivative of f(x)=sqrt(x)(5e^x-4x^4) ?
The first derivative of f(x)=sqrt(x)(5e^x-4x^4) is (-36x^4+10e^xx+5e^x)/(2sqrt(x))