What is the derivative of v^{3/2}(v+ce^v) ?

The derivative of v^{3/2}(v+ce^v) is (3ce^vv^{1/2}+2ce^vv^{3/2}+5v^{3/2})/2

What is the first derivative of v^{3/2}(v+ce^v) ?

The first derivative of v^{3/2}(v+ce^v) is (3ce^vv^{1/2}+2ce^vv^{3/2}+5v^{3/2})/2

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derivative of

v^{\frac{3}{2}} (v + c e^{v})

\frac{3 c e ^{v} v ^{\frac{1}{2}} + 2 c e ^{v} v ^{\frac{3}{2}} + 5 v ^{\frac{3}{2}}}{2}

Solution steps

\frac{d}{d v} (v^{\frac{3}{2}} (v + c e^{v}))

Apply the Product Rule:

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

f = v^{\frac{3}{2}}, g = v + c e^{v}

= \frac{d}{d v} (v^{\frac{3}{2}}) (v + c e^{v}) + \frac{d}{d v} (v + c e^{v}) v^{\frac{3}{2}}

\frac{d}{d v} (v^{\frac{3}{2}}) = \frac{3 v ^{\frac{1}{2}}}{2}

\frac{d}{d v} (v + c e^{v}) = 1 + c e^{v}

= \frac{3 v ^{\frac{1}{2}}}{2} (v + c e^{v}) + (1 + c e^{v}) v^{\frac{3}{2}}

Simplify

\frac{3 v ^{\frac{1}{2}}}{2} (v + c e^{v}) + (1 + c e^{v}) v^{\frac{3}{2}} : \frac{3 c e ^{v} v ^{\frac{1}{2}} + 2 c e ^{v} v ^{\frac{3}{2}} + 5 v ^{\frac{3}{2}}}{2}

= \frac{3 c e ^{v} v ^{\frac{1}{2}} + 2 c e ^{v} v ^{\frac{3}{2}} + 5 v ^{\frac{3}{2}}}{2}

\frac{d}{d x} (\frac{5 x - 2}{4 x + 3})

f (x) = arcsec (e^{2 x})

y^{^{'}} + \frac{1}{2} y = 5

x \to 7 - lim (x - 7)

\int sin (5 x) ln (sin (5 x)) d x

What is the derivative of v^{3/2}(v+ce^v) ?
The derivative of v^{3/2}(v+ce^v) is (3ce^vv^{1/2}+2ce^vv^{3/2}+5v^{3/2})/2
What is the first derivative of v^{3/2}(v+ce^v) ?
The first derivative of v^{3/2}(v+ce^v) is (3ce^vv^{1/2}+2ce^vv^{3/2}+5v^{3/2})/2