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

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

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

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

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derivative of

- \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}

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