What is the derivative of (2x+e^x(5-sqrt(x))) ?

The derivative of (2x+e^x(5-sqrt(x))) is (-6x-2e^xx+10e^xsqrt(x)-e^x)/(2sqrt(x))+10

What is the first derivative of (2x+e^x(5-sqrt(x))) ?

The first derivative of (2x+e^x(5-sqrt(x))) is (-6x-2e^xx+10e^xsqrt(x)-e^x)/(2sqrt(x))+10

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

\frac{- 6 x - 2 e ^{x} x + 10 e ^{x} x - e ^{x}}{2 x} + 10

Solution steps

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

Apply the Product Rule:

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

f = 2 x + e^{x}, g = 5 - x

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

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

\frac{d}{d x} (5 - x) = - \frac{1}{2 x}

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

Simplify

(2 + e^{x}) (5 - x) + (- \frac{1}{2 x}) (2 x + e^{x}) : \frac{- 6 x - 2 e ^{x} x + 10 e ^{x} x - e ^{x}}{2 x} + 10

= \frac{- 6 x - 2 e ^{x} x + 10 e ^{x} x - e ^{x}}{2 x} + 10

\frac{2 x ^{2}}{x ^{2} - 1}

\frac{\partial}{\partial x} (e^{x^{2} y^{4}} sin (x yz))

4 (e^{x})^{9}

y^{^{''}} - y^{^{'}} = x

\frac{d}{d x} ((2 x)^{- 4} - (2 t^{- 1})^{2})

What is the derivative of (2x+e^x(5-sqrt(x))) ?
The derivative of (2x+e^x(5-sqrt(x))) is (-6x-2e^xx+10e^xsqrt(x)-e^x)/(2sqrt(x))+10
What is the first derivative of (2x+e^x(5-sqrt(x))) ?
The first derivative of (2x+e^x(5-sqrt(x))) is (-6x-2e^xx+10e^xsqrt(x)-e^x)/(2sqrt(x))+10