How do you differentiate $y = 3^{cot x}$ $y=3^(cotx)$ ?

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Answer 1 · 2016-07-21T00:21:00

This is an interesting problem.

There are different strategies of approaching this question, but I will take what I see simplest: use of the chain rule.

Let $y = 3^{u}$ $y = 3^u$ and $u = cot x$ $u = cotx$ . We have to differentiate both.

$y = 3^{u}$ $y = 3^u$

$ln (y) = ln (3^{u})$ $ln(y) = ln(3^u)$

$ln y = u (ln 3)$ $lny = u(ln3)$

$\frac{1}{y} (\frac{d y}{d u}) = ln 3$ $1/y(dy/(du)) = ln3$

$\frac{d y}{d} u = y ln 3$ $dy/du = yln3$

$\frac{d y}{d u} = 3^{u} ln 3$ $dy/(du) = 3^(u)ln3$

Now for $u$ $u$ :

$u = \frac{cos x}{sin x}$ $u = cosx/sinx$

$u' = (-sinx xx sinx - cosx xx cosx)/(sinx)^2$

$u' = (-sin^2x - cos^2x)/(sin^2x)$

$u' = (-(sin^2x + cos^2x))/(sin^2x)$

$u' = (-1)/sin^2x$

$u' = -csc^2x$

By the chain rule:

$dy/dx = 3^(u)ln3 xx -csc^2x$

$dy/dx = -csc^2x xx 3^(cotx)xxln3$

In summary, the derivative of $y = 3^(cot(x))$ is $dy/dx = -csc^2x xx 3^(cotx)xxln3$ .

Hopefully this helps!

How do you differentiate y=3^(cotx)y=3cotxy=3^(cotx)?