en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of y=x^{1-x}

Solution

derivative of

y = x^{1 - x}

Solution

x^{1 - x} (- ln (x) + \frac{1 - x}{x})

Solution steps

\frac{d}{d x} (x^{1 - x})

Apply exponent rule:

a^{b} = e^{b l n (a)}

x^{1 - x} = e^{(1 - x) l n (x)}

= \frac{d}{d x} (e^{(1 - x) l n (x)})

Apply the chain rule:

e^{(1 - x) l n (x)} \frac{d}{d x} ((1 - x) ln (x))

= e^{(1 - x) l n (x)} \frac{d}{d x} ((1 - x) ln (x))

\frac{d}{d x} ((1 - x) ln (x)) = - ln (x) + \frac{1 - x}{x}

= e^{(1 - x) l n (x)} (- ln (x) + \frac{1 - x}{x})

e^{(1 - x) l n (x)} = x^{1 - x}

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

Popular Examples

tangent of f(x)= 1/((1+x^2))tangent of

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

derivative of 4-(20/(x^3)+3/(x^5))

\frac{d}{d x} (4 - \frac{20}{x ^{3}} + \frac{3}{x ^{5}})

tangent of f(x)=x^2+x,\at (x=1)tangent of

f (x) = x^{2} + x, at (x = 1)

(\partial)/(\partial x)(y^2+x^2)

\frac{\partial}{\partial x} (y^{2} + x^{2})

3xy^2y^'=3x^4+y^3

3 x y^{2} y^{^{'}} = 3 x^{4} + y^{3}

Frequently Asked Questions (FAQ)

What is the derivative of y=x^{1-x} ?
The derivative of y=x^{1-x} is x^{1-x}(-ln(x)+(1-x)/x)
What is the first derivative of y=x^{1-x} ?
The first derivative of y=x^{1-x} is x^{1-x}(-ln(x)+(1-x)/x)