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))

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\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 )}

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))