What is the derivative of (r^6-9r)e^r ?

The derivative of (r^6-9r)e^r is (6r^5-9)e^r+e^r(r^6-9r)

What is the first derivative of (r^6-9r)e^r ?

The first derivative of (r^6-9r)e^r is (6r^5-9)e^r+e^r(r^6-9r)

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derivative of

(r^{6} - 9 r) e^{r}

(6 r^{5} - 9) e^{r} + e^{r} (r^{6} - 9 r)

Solution steps

\frac{d}{d r} ((r^{6} - 9 r) e^{r})

Apply the Product Rule:

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

f = r^{6} - 9 r, g = e^{r}

= \frac{d}{d r} (r^{6} - 9 r) e^{r} + \frac{d}{d r} (e^{r}) (r^{6} - 9 r)

\frac{d}{d r} (r^{6} - 9 r) = 6 r^{5} - 9

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

= (6 r^{5} - 9) e^{r} + e^{r} (r^{6} - 9 r)

\int \frac{8 x}{x - 1} d x

f (x) = 6 x^{3}

\frac{d}{d z} (\frac{z}{( z ^{2} + R ^{2} ) ^{\frac{3}{2}}})

\int \frac{0}{x} d x

\frac{d}{d x} (\frac{e ^{x}}{x ^{7}})

What is the derivative of (r^6-9r)e^r ?
The derivative of (r^6-9r)e^r is (6r^5-9)e^r+e^r(r^6-9r)
What is the first derivative of (r^6-9r)e^r ?
The first derivative of (r^6-9r)e^r is (6r^5-9)e^r+e^r(r^6-9r)