What is the derivative of f(x)=5e^xsqrt(x) ?

The derivative of f(x)=5e^xsqrt(x) is 5(e^xsqrt(x)+(e^x)/(2sqrt(x)))

What is the first derivative of f(x)=5e^xsqrt(x) ?

The first derivative of f(x)=5e^xsqrt(x) is 5(e^xsqrt(x)+(e^x)/(2sqrt(x)))

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derivative of

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

5 (e^{x} x + \frac{e ^{x}}{2 x})

Solution steps

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

Take the constant out:

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

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

Apply the Product Rule:

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

f = e^{x}, g = x

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

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

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

= 5 (e^{x} x + \frac{1}{2 x} e^{x})

\frac{1}{2 x} e^{x} = \frac{e ^{x}}{2 x}

= 5 (e^{x} x + \frac{e ^{x}}{2 x})

\int \frac{36 - 25 x ^{2}}{x} d x

x \to - 6 lim (\frac{- 8 x}{x + 6})

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

y^{^{'}} + y (1 - 2 x) = - \frac{x ^{2}}{2} y

y = e^{\frac{1}{x ^{2}}} + \frac{1}{e ^{x^{2}}}

What is the derivative of f(x)=5e^xsqrt(x) ?
The derivative of f(x)=5e^xsqrt(x) is 5(e^xsqrt(x)+(e^x)/(2sqrt(x)))
What is the first derivative of f(x)=5e^xsqrt(x) ?
The first derivative of f(x)=5e^xsqrt(x) is 5(e^xsqrt(x)+(e^x)/(2sqrt(x)))