What is the derivative of y=sqrt(x)(x^4-x^3) ?

The derivative of y=sqrt(x)(x^4-x^3) is (x^{5/2}(x-1))/2+sqrt(x)(4x^3-3x^2)

What is the first derivative of y=sqrt(x)(x^4-x^3) ?

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

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derivative of

y = x (x^{4} - x^{3})

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

Solution steps

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

Apply the Product Rule:

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

f = x, g = x^{4} - x^{3}

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

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

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

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

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

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

y = \frac{36 x}{x ^{2} + 36}, (- 3, - \frac{12}{5})

\frac{d}{d x} (21 x + x^{2})

\frac{d}{d x} (e^{2 x})

\frac{\partial}{\partial x} (6 x + y^{3})

\frac{d}{d x} (x^{3} - 6 x^{2} + 12 x)

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