What is the derivative of (x^{1/2}*(x^2+x-4)) ?

The derivative of (x^{1/2}*(x^2+x-4)) is (5x^2+3x-4)/(2x^{1/2)}

What is the first derivative of (x^{1/2}*(x^2+x-4)) ?

The first derivative of (x^{1/2}*(x^2+x-4)) is (5x^2+3x-4)/(2x^{1/2)}

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

\frac{5 x ^{2} + 3 x - 4}{2 x ^{\frac{1}{2}}}

Solution steps

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

Apply the Product Rule:

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

f = x^{\frac{1}{2}}, g = x^{2} + x - 4

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

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

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

= \frac{1}{2 x ^{\frac{1}{2}}} (x^{2} + x - 4) + (2 x + 1) x^{\frac{1}{2}}

Simplify

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

= \frac{5 x ^{2} + 3 x - 4}{2 x ^{\frac{1}{2}}}

\frac{d}{d x} (x^{3} + b x^{2} + c x + d)

\frac{\partial}{\partial y} (sin (x y) + x sin (y))

n = 0 \sum \infty (\frac{5}{5})^{n}

x y^{^{'}} + 5 y = x^{3} - x

\frac{\partial}{\partial s} (s \cdot cos (t))

What is the derivative of (x^{1/2}*(x^2+x-4)) ?
The derivative of (x^{1/2}*(x^2+x-4)) is (5x^2+3x-4)/(2x^{1/2)}
What is the first derivative of (x^{1/2}*(x^2+x-4)) ?
The first derivative of (x^{1/2}*(x^2+x-4)) is (5x^2+3x-4)/(2x^{1/2)}