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})

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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})

x \to 0 + lim (x^{x^{x}})

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

n = 0 \sum \infty \frac{1}{n ^{2} + 4 n + 3}

y = 2 x^{3} - x^{2} + 5, (2, 17)

f (x) = 3 x^{2} + 4 x + 1, at x = 0

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})