en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of 2/(\sqrt[4]{x^3})

Solution

Solution

- \frac{3}{2 x ^{\frac{7}{4}}}

Solution steps

Take the constant out:

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

Apply exponent rule:

\frac{1}{a} = a^{- 1}

= 2 \frac{d}{d x} ((x^{3})^{- \frac{1}{4}})

Apply the chain rule:

- \frac{1}{4 ( x ^{3} ) ^{\frac{5}{4}}} \frac{d}{d x} (x^{3})

= - \frac{1}{4 ( x ^{3} ) ^{\frac{5}{4}}} \frac{d}{d x} (x^{3})

\frac{d}{d x} (x^{3}) = 3 x^{2}

= 2 (- \frac{1}{4 ( x ^{3} ) ^{\frac{5}{4}}} \cdot 3 x^{2})

Simplify

2 (- \frac{1}{4 ( x ^{3} ) ^{\frac{5}{4}}} \cdot 3 x^{2}) : - \frac{3}{2 x ^{\frac{7}{4}}}

= - \frac{3}{2 x ^{\frac{7}{4}}}

Popular Examples

sqrt(x)(dy)/(dx)=e^{y+sqrt(x)}

x \frac{d y}{d x} = e^{y + x}

tangent of f(x)=x-x^3,(1,0)tangent of

f (x) = x - x^{3}, (1, 0)

tangent of x^2+2xy-y^2+x=27,(4,7)tangent of

x^{2} + 2 x y - y^{2} + x = 27, (4, 7)

(dy)/(dx)=(y^3)/(1-2xy^2),y(0)=1

\frac{d y}{d x} = \frac{y ^{3}}{1 - 2 x y ^{2}}, y (0) = 1

(\partial)/(\partial x)(sin(z^2)cos(xy))

\frac{\partial}{\partial x} (sin (z^{2}) cos (x y))

Frequently Asked Questions (FAQ)

What is the derivative of 2/(\sqrt[4]{x^3}) ?
The derivative of 2/(\sqrt[4]{x^3}) is -3/(2x^{7/4)}
What is the first derivative of 2/(\sqrt[4]{x^3}) ?
The first derivative of 2/(\sqrt[4]{x^3}) is -3/(2x^{7/4)}