What is the derivative of y^{ln(x}) ?

The derivative of y^{ln(x}) is x^{ln(y)}((ln(y))/x+(ln(x)(dy)/(dx))/y)

What is the first derivative of y^{ln(x}) ?

The first derivative of y^{ln(x}) is x^{ln(y)}((ln(y))/x+(ln(x)(dy)/(dx))/y)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

\frac{d}{d x} (y^{l n (x)})

x^{l n (y)} (\frac{ln ( y )}{x} + \frac{ln ( x ) \frac{d y}{d x}}{y})

Solution steps

\frac{d}{d x} (y^{l n (x)})

Apply exponent rule:

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

y^{l n (x)} = e^{l n (x) l n (y)}

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

Apply the chain rule:

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

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

\frac{d}{d x} (ln (x) ln (y)) = \frac{ln ( y )}{x} + \frac{ln ( x ) \frac{d y}{d x}}{y}

= e^{l n (x) l n (y)} (\frac{ln ( y )}{x} + \frac{ln ( x ) \frac{d y}{d x}}{y})

e^{l n (x) l n (y)} = x^{l n (y)}

= x^{l n (y)} (\frac{ln ( y )}{x} + \frac{ln ( x ) \frac{d y}{d x}}{y})

(2.52 \cdot 1 0^{- 5}) \cdot 362226.816

20 \cdot e^{- 0.8 (3.8)}

200 \cdot e^{- 0.086 \cdot 30}

e^{- 1} \cdot ln (e^{- 1})

4.9 \cdot 1 0^{- 7}

What is the derivative of y^{ln(x}) ?
The derivative of y^{ln(x}) is x^{ln(y)}((ln(y))/x+(ln(x)(dy)/(dx))/y)
What is the first derivative of y^{ln(x}) ?
The first derivative of y^{ln(x}) is x^{ln(y)}((ln(y))/x+(ln(x)(dy)/(dx))/y)