What is the derivative of 7x^{1/5}(x^3-2) ?

The derivative of 7x^{1/5}(x^3-2) is (7(16x^3-2))/(5x^{4/5)}

What is the first derivative of 7x^{1/5}(x^3-2) ?

The first derivative of 7x^{1/5}(x^3-2) is (7(16x^3-2))/(5x^{4/5)}

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derivative of

7 x^{\frac{1}{5}} (x^{3} - 2)

\frac{7 ( 16 x ^{3} - 2 )}{5 x ^{\frac{4}{5}}}

Solution steps

\frac{d}{d x} (7 x^{\frac{1}{5}} (x^{3} - 2))

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 7 \frac{d}{d x} (x^{\frac{1}{5}} (x^{3} - 2))

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x^{\frac{1}{5}}, g = x^{3} - 2

= 7 (\frac{d}{d x} (x^{\frac{1}{5}}) (x^{3} - 2) + \frac{d}{d x} (x^{3} - 2) x^{\frac{1}{5}})

\frac{d}{d x} (x^{\frac{1}{5}}) = \frac{1}{5 x ^{\frac{4}{5}}}

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

= 7 (\frac{1}{5 x ^{\frac{4}{5}}} (x^{3} - 2) + 3 x^{2} x^{\frac{1}{5}})

Simplify

7 (\frac{1}{5 x ^{\frac{4}{5}}} (x^{3} - 2) + 3 x^{2} x^{\frac{1}{5}}) : \frac{7 ( 16 x ^{3} - 2 )}{5 x ^{\frac{4}{5}}}

= \frac{7 ( 16 x ^{3} - 2 )}{5 x ^{\frac{4}{5}}}

\frac{d}{d x} (y^{2} (8 - x))

x^{2} y^{^{'}} = 4 x^{2} y^{3} - 2 x y

n = 1 \sum \infty \frac{( 3 ^{n} )}{n !}

cos (4 t) cos (6 t)

\int 5 sin^{3} (x) cos (x) d x

What is the derivative of 7x^{1/5}(x^3-2) ?
The derivative of 7x^{1/5}(x^3-2) is (7(16x^3-2))/(5x^{4/5)}
What is the first derivative of 7x^{1/5}(x^3-2) ?
The first derivative of 7x^{1/5}(x^3-2) is (7(16x^3-2))/(5x^{4/5)}