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Popular >

derivative of (r^5-4r)e^r

Solution

derivative of

(r^{5} - 4 r) e^{r}

Solution

(5 r^{4} - 4) e^{r} + e^{r} (r^{5} - 4 r)

Solution steps

\frac{d}{d r} ((r^{5} - 4 r) e^{r})

Apply the Product Rule:

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

f = r^{5} - 4 r, g = e^{r}

= \frac{d}{d r} (r^{5} - 4 r) e^{r} + \frac{d}{d r} (e^{r}) (r^{5} - 4 r)

\frac{d}{d r} (r^{5} - 4 r) = 5 r^{4} - 4

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

= (5 r^{4} - 4) e^{r} + e^{r} (r^{5} - 4 r)

Popular Examples

(dy)/(dx)=(sqrt(y^2-9)),y(1)=4

\frac{d y}{d x} = (y^{2} - 9), y (1) = 4

integral of e^xcos(x/2)

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

integral of 2/(x^{-1)+1}

\int \frac{2}{x ^{- 1} + 1} d x

integral of 4/(x^2-4x+9)

\int \frac{4}{x ^{2} - 4 x + 9} d x

deg2rad ((x')/x)^'degrees to radians

(\frac{x ^{'}}{x})^{^{'}}

Frequently Asked Questions (FAQ)

What is the derivative of (r^5-4r)e^r ?
The derivative of (r^5-4r)e^r is (5r^4-4)e^r+e^r(r^5-4r)
What is the first derivative of (r^5-4r)e^r ?
The first derivative of (r^5-4r)e^r is (5r^4-4)e^r+e^r(r^5-4r)