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

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

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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

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

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

Solution steps

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

Treat

c, x

as a constant

Apply the Product Rule:

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

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

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

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

Simplify

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

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

(y + x cos^{2} (\frac{y}{x})) d x - x d y = 0

\int \frac{6 x}{1 + x ^{4}} d x

\frac{d x}{d t} = 349.344897 (1 - e^{- 0.0916 t})

\int (3 e^{- 0.4 x}) d x

\int 4 sin (x) - 1 + 8 x^{- 5} d x

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