What is the derivative of \sqrt[3]{5x}+4x^{1/2}-3x ?

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

What is the first derivative of \sqrt[3]{5x}+4x^{1/2}-3x ?

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

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

\frac{5 ^{\frac{1}{3}}}{3 x ^{\frac{2}{3}}} + \frac{2}{x ^{\frac{1}{2}}} - 3

Solution steps

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

Apply the Sum/Difference Rule:

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

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

\frac{d}{d x} (3 5 x) = \frac{5 ^{\frac{1}{3}}}{3 x ^{\frac{2}{3}}}

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

\frac{d}{d x} (3 x) = 3

= \frac{5 ^{\frac{1}{3}}}{3 x ^{\frac{2}{3}}} + \frac{2}{x ^{\frac{1}{2}}} - 3

x^{2} \frac{d y}{d x} - 2 x y = 6 y^{4}, y (1) = \frac{1}{5}

\frac{d}{d x} (2 x ln (4 x^{2}))

\int \frac{60}{2 x + 5} - 5 d x

\int cot (x) (- 8 sin (x) - 10 csc (x)) d x

y^{2} + x^{2} y^{^{'}} = 0

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