What is the derivative of (2ra^{rx}+n)^p ?

The derivative of (2ra^{rx}+n)^p is p(2ra^{rx}+n)^{p-1}*2(xa^{xr}rln(a)+a^{xr})

What is the first derivative of (2ra^{rx}+n)^p ?

The first derivative of (2ra^{rx}+n)^p is p(2ra^{rx}+n)^{p-1}*2(xa^{xr}rln(a)+a^{xr})

derivative of

(2 r a^{r x} + n)^{p}

p (2 r a^{r x} + n)^{p - 1} \cdot 2 (x a^{x r} r ln (a) + a^{x r})

Solution steps

\frac{d}{d r} ((2 r a^{r x} + n)^{p})

Treat

a, n, x, p

as constants

Apply the chain rule:

p (2 r a^{r x} + n)^{p - 1} \frac{d}{d r} (2 r a^{r x} + n)

= p (2 r a^{r x} + n)^{p - 1} \frac{d}{d r} (2 r a^{r x} + n)

\frac{d}{d r} (2 r a^{r x} + n) = 2 (x a^{x r} r ln (a) + a^{x r})

= p (2 r a^{r x} + n)^{p - 1} \cdot 2 (x a^{x r} r ln (a) + a^{x r})

\frac{d}{d x} (x - 0.8 - 0.2 sin (x))

8 cos^{2} (y) d x + csc^{2} (x) d y = 0, y (\frac{π}{12}) = \frac{π}{4}

\frac{d}{d x} (\frac{a + b x + c x ^{2}}{x})

\frac{d y}{d x} + 4 y = e^{- 4 x}

\frac{\partial}{\partial x} (3 x^{2} - 3)

What is the derivative of (2ra^{rx}+n)^p ?
The derivative of (2ra^{rx}+n)^p is p(2ra^{rx}+n)^{p-1}*2(xa^{xr}rln(a)+a^{xr})
What is the first derivative of (2ra^{rx}+n)^p ?
The first derivative of (2ra^{rx}+n)^p is p(2ra^{rx}+n)^{p-1}*2(xa^{xr}rln(a)+a^{xr})