What is the derivative of kx^4(x_{0}-x) ?

The derivative of kx^4(x_{0}-x) is k(4x^3x_{0}-5x^4)

What is the first derivative of kx^4(x_{0}-x) ?

The first derivative of kx^4(x_{0}-x) is k(4x^3x_{0}-5x^4)

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\frac{d}{d x} (k x^{4} (x_{0} - x))

k (4 x^{3} x_{0} - 5 x^{4})

Solution steps

\frac{d}{d x} (k x^{4} (x_{0} - x))

Treat

k, x_{0}

as a constant

Take the constant out:

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

= k \frac{d}{d x} (x^{4} (x_{0} - x))

Apply the Product Rule:

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

f = x^{4}, g = x_{0} - x

= k (\frac{d}{d x} (x^{4}) (x_{0} - x) + \frac{d}{d x} (x_{0} - x) x^{4})

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

\frac{d}{d x} (x_{0} - x) = - 1

= k (4 x^{3} (x_{0} - x) + (- 1) x^{4})

Simplify

k (4 x^{3} (x_{0} - x) + (- 1) x^{4}) : k (4 x^{3} x_{0} - 5 x^{4})

= k (4 x^{3} x_{0} - 5 x^{4})

\frac{d}{d x} ((sin (5 x^{2}))^{2})

\int_{1}^{5} \frac{x - 1}{x ^{3} + x} d x

9 y^{^{'}} + y = 5 t^{2}

n = 1 \sum \infty (- \frac{3}{5})^{n}

\frac{d}{d u} (tan (\frac{u}{v}))

What is the derivative of kx^4(x_{0}-x) ?
The derivative of kx^4(x_{0}-x) is k(4x^3x_{0}-5x^4)
What is the first derivative of kx^4(x_{0}-x) ?
The first derivative of kx^4(x_{0}-x) is k(4x^3x_{0}-5x^4)