What is the derivative of h(x)=(4x+6)^x ?

The derivative of h(x)=(4x+6)^x is (ln(4x+6)+(2x)/(2x+3))(4x+6)^x

What is the first derivative of h(x)=(4x+6)^x ?

The first derivative of h(x)=(4x+6)^x is (ln(4x+6)+(2x)/(2x+3))(4x+6)^x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

h (x) = (4 x + 6)^{x}

(ln (4 x + 6) + \frac{2 x}{2 x + 3}) (4 x + 6)^{x}

Solution steps

\frac{d}{d x} ((4 x + 6)^{x})

Apply exponent rule:

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

(4 x + 6)^{x} = e^{x l n (4 x + 6)}

= \frac{d}{d x} (e^{x l n (4 x + 6)})

Apply the chain rule:

e^{x l n (4 x + 6)} \frac{d}{d x} (x ln (4 x + 6))

= e^{x l n (4 x + 6)} \frac{d}{d x} (x ln (4 x + 6))

\frac{d}{d x} (x ln (4 x + 6)) = ln (4 x + 6) + \frac{2 x}{2 x + 3}

= e^{x l n (4 x + 6)} (ln (4 x + 6) + \frac{2 x}{2 x + 3})

e^{x l n (4 x + 6)} = (4 x + 6)^{x}

= (ln (4 x + 6) + \frac{2 x}{2 x + 3}) (4 x + 6)^{x}

x \to - 2 lim (3^{x^{2 + 1}})

\int_{0}^{4} \frac{1}{x ^{2} - 4 x - 5} d x

\frac{d}{d x} (8 x^{\frac{2}{3}} + x^{\frac{5}{3}})

x \to - 4 lim (\frac{1}{4} + \frac{1}{x})

\frac{d}{d x} ((cos (6 x))^{x})

What is the derivative of h(x)=(4x+6)^x ?
The derivative of h(x)=(4x+6)^x is (ln(4x+6)+(2x)/(2x+3))(4x+6)^x
What is the first derivative of h(x)=(4x+6)^x ?
The first derivative of h(x)=(4x+6)^x is (ln(4x+6)+(2x)/(2x+3))(4x+6)^x