What is the derivative of 5/(\sqrt[3]{x}-6/(2^x)) ?

The derivative of 5/(\sqrt[3]{x}-6/(2^x)) is -5/(3x^{4/3)}+3ln(2)*2^{1-x}

What is the first derivative of 5/(\sqrt[3]{x}-6/(2^x)) ?

The first derivative of 5/(\sqrt[3]{x}-6/(2^x)) is -5/(3x^{4/3)}+3ln(2)*2^{1-x}

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- \frac{5}{3 x ^{\frac{4}{3}}} + 3 ln (2) \cdot 2^{1 - x}

Solution steps

Simplify

Apply the Sum/Difference Rule:

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

\frac{d}{d x} (3 \cdot 2^{1 - x}) = - 3 ln (2) \cdot 2^{1 - x}

= - \frac{5}{3 x ^{\frac{4}{3}}} - (- 3 ln (2) \cdot 2^{1 - x})

Simplify

= - \frac{5}{3 x ^{\frac{4}{3}}} + 3 ln (2) \cdot 2^{1 - x}

y^{^{'}} - \frac{3}{2} y = 3 t + 2 e^{t}

- 14 e^{- x^{2}} x

3 (y^{4} + 1) d x + (4 x y^{3}) d y = 0

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

\int \frac{1}{( y - 1 )} d y

What is the derivative of 5/(\sqrt[3]{x}-6/(2^x)) ?
The derivative of 5/(\sqrt[3]{x}-6/(2^x)) is -5/(3x^{4/3)}+3ln(2)*2^{1-x}
What is the first derivative of 5/(\sqrt[3]{x}-6/(2^x)) ?
The first derivative of 5/(\sqrt[3]{x}-6/(2^x)) is -5/(3x^{4/3)}+3ln(2)*2^{1-x}