What is the derivative of 6e^x+8/(\sqrt[3]{x}) ?

The derivative of 6e^x+8/(\sqrt[3]{x}) is 6e^x-8/(3x^{4/3)}

What is the first derivative of 6e^x+8/(\sqrt[3]{x}) ?

The first derivative of 6e^x+8/(\sqrt[3]{x}) is 6e^x-8/(3x^{4/3)}

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6 e^{x} - \frac{8}{3 x ^{\frac{4}{3}}}

Solution steps

Apply the Sum/Difference Rule:

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

\frac{d}{d x} (6 e^{x}) = 6 e^{x}

= 6 e^{x} - \frac{8}{3 x ^{\frac{4}{3}}}

\frac{d y}{d x} = \frac{( x ^{2} )}{y ( 1 + x ^{3} )}

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

\frac{d}{d x} (cos^{4} (x) - sin^{4} (x))

k = 1 \sum \infty a (k - 1) x^{k}

What is the derivative of 6e^x+8/(\sqrt[3]{x}) ?
The derivative of 6e^x+8/(\sqrt[3]{x}) is 6e^x-8/(3x^{4/3)}
What is the first derivative of 6e^x+8/(\sqrt[3]{x}) ?
The first derivative of 6e^x+8/(\sqrt[3]{x}) is 6e^x-8/(3x^{4/3)}