What is the derivative of 2(3^{-x}^2+log_{2}(x^4-1)) ?

The derivative of 2(3^{-x}^2+log_{2}(x^4-1)) is -4ln(3)*9^{-x}+(4x^3)/(ln(2)(x^4-1))

What is the first derivative of 2(3^{-x}^2+log_{2}(x^4-1)) ?

The first derivative of 2(3^{-x}^2+log_{2}(x^4-1)) is -4ln(3)*9^{-x}+(4x^3)/(ln(2)(x^4-1))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

\frac{d}{d x} (2 (3^{- x})^{2} + lo g_{2} (x^{4} - 1))

- 4 ln (3) \cdot 9^{- x} + \frac{4 x ^{3}}{ln ( 2 ) ( x ^{4} - 1 )}

Solution steps

\frac{d}{d x} (2 (3^{- x})^{2} + lo g_{2} (x^{4} - 1))

Simplify

2 (3^{- x})^{2} + lo g_{2} (x^{4} - 1) : 2 \cdot 9^{- x} + lo g_{2} (x^{4} - 1)

= \frac{d}{d x} (2 \cdot 9^{- x} + lo g_{2} (x^{4} - 1))

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (2 \cdot 9^{- x}) + \frac{d}{d x} (lo g_{2} (x^{4} - 1))

\frac{d}{d x} (2 \cdot 9^{- x}) = - 4 ln (3) \cdot 9^{- x}

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

= - 4 ln (3) \cdot 9^{- x} + \frac{4 x ^{3}}{ln ( 2 ) ( x ^{4} - 1 )}

e^{1 - 2 t} + cos^{2} (t)

f (x) = \frac{18 x + 8 x ^{2}}{( 9 + 8 x ) ^{2}}

\int x^{3} x^{2} - 25 d x

f (x) = e^{x + 1} - 3 e^{x} + 2 e^{x^{3}}

\int_{2}^{\infty} 2 x^{- 2} d x

What is the derivative of 2(3^{-x}^2+log_{2}(x^4-1)) ?
The derivative of 2(3^{-x}^2+log_{2}(x^4-1)) is -4ln(3)*9^{-x}+(4x^3)/(ln(2)(x^4-1))
What is the first derivative of 2(3^{-x}^2+log_{2}(x^4-1)) ?
The first derivative of 2(3^{-x}^2+log_{2}(x^4-1)) is -4ln(3)*9^{-x}+(4x^3)/(ln(2)(x^4-1))