What is the derivative of (e^x(5e^{5x}-25x)) ?

The derivative of (e^x(5e^{5x}-25x)) is -25e^xx+30e^{6x}-25e^x

What is the first derivative of (e^x(5e^{5x}-25x)) ?

The first derivative of (e^x(5e^{5x}-25x)) is -25e^xx+30e^{6x}-25e^x

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\frac{d}{d x} ((e^{x}) (5 e^{5 x} - 25 x))

- 25 e^{x} x + 30 e^{6 x} - 25 e^{x}

Solution steps

\frac{d}{d x} ((e^{x}) (5 e^{5 x} - 25 x))

Apply the Product Rule:

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

f = e^{x}, g = 5 e^{5 x} - 25 x

= \frac{d}{d x} (e^{x}) (5 e^{5 x} - 25 x) + \frac{d}{d x} (5 e^{5 x} - 25 x) e^{x}

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

\frac{d}{d x} (5 e^{5 x} - 25 x) = 25 e^{5 x} - 25

= e^{x} (5 e^{5 x} - 25 x) + (25 e^{5 x} - 25) e^{x}

Simplify

e^{x} (5 e^{5 x} - 25 x) + (25 e^{5 x} - 25) e^{x} : - 25 e^{x} x + 30 e^{6 x} - 25 e^{x}

= - 25 e^{x} x + 30 e^{6 x} - 25 e^{x}

(4 x - 8 y = - 1, 4) (2 x - 4 y = - 7)

(2 x - 6 x y^{3}) (e^{t})

2 l e f t (3, 4, 5) - 3 l e f t (- 1, - 2, - 3)

\frac{( x ^{2} + 1 ) ^{2}}{2}

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

What is the derivative of (e^x(5e^{5x}-25x)) ?
The derivative of (e^x(5e^{5x}-25x)) is -25e^xx+30e^{6x}-25e^x
What is the first derivative of (e^x(5e^{5x}-25x)) ?
The first derivative of (e^x(5e^{5x}-25x)) is -25e^xx+30e^{6x}-25e^x