What is the derivative of 3x^{1/4}(x^3-5) ?

The derivative of 3x^{1/4}(x^3-5) is (3(13x^3-5))/(4x^{3/4)}

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

The first derivative of 3x^{1/4}(x^3-5) is (3(13x^3-5))/(4x^{3/4)}

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derivative of

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

\frac{3 ( 13 x ^{3} - 5 )}{4 x ^{\frac{3}{4}}}

Solution steps

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

Take the constant out:

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

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

Apply the Product Rule:

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

f = x^{\frac{1}{4}}, g = x^{3} - 5

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

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

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

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

Simplify

3 (\frac{1}{4 x ^{\frac{3}{4}}} (x^{3} - 5) + 3 x^{2} x^{\frac{1}{4}}) : \frac{3 ( 13 x ^{3} - 5 )}{4 x ^{\frac{3}{4}}}

= \frac{3 ( 13 x ^{3} - 5 )}{4 x ^{\frac{3}{4}}}

\int_{0}^{\infty} sin (5 x) d x

\frac{\partial}{\partial x} (e^{2 x} y)

\frac{\partial}{\partial x} (x^{2} + 4)

y = 8 x^{\frac{1}{2}} + 4 x^{\frac{3}{2}} + 4

x \to \infty lim (\frac{x - cos ( x )}{x})

What is the derivative of 3x^{1/4}(x^3-5) ?
The derivative of 3x^{1/4}(x^3-5) is (3(13x^3-5))/(4x^{3/4)}
What is the first derivative of 3x^{1/4}(x^3-5) ?
The first derivative of 3x^{1/4}(x^3-5) is (3(13x^3-5))/(4x^{3/4)}