# What are the derivatives of these functions? log_4(6x - 4), log_3(x^2/(x-4)), xcdot4^(-2x), "arc" cos(4sqrtx), xarctan(4x-6)

May 29, 2015

There are certain derivatives you'll just have to memorize. For these problems, we have to know:

$\frac{d}{\mathrm{dx}} \left({\log}_{10} x\right) = \frac{d}{\mathrm{dx}} \left(\ln \frac{x}{\ln 10}\right) = \frac{1}{x \cdot \ln 10}$
(could be derived from the change of base law!)

$\frac{d}{\mathrm{dx}} {a}^{b x} = \frac{d}{\mathrm{dx}} {\left({a}^{x}\right)}^{b} = b \cdot {\left({a}^{x}\right)}^{b - 1} \cdot \frac{d}{\mathrm{dx}} \left({a}^{x}\right)$

$\frac{d}{\mathrm{dx}} \left({a}^{x}\right) = {a}^{x} \ln a$
(must memorize this one)

$\frac{d}{\mathrm{dx}} {a}^{b x} = b \cdot {\left({a}^{x}\right)}^{b - 1} \cdot {a}^{x} \ln a = b {a}^{b x - x} \cdot {a}^{x} \ln a$

$= b {a}^{b x} \ln a$
(can be derived)

Then there are all those inverse trig derivatives. Geez.

d/(dx)(arcsinx")" = 1/(sqrt(1-x^2)
d/(dx)(arccosx")" = -1/(sqrt(1-x^2)
d/(dx)(arc$\tan x \text{)} = \frac{1}{1 + {x}^{2}}$
d/(dx)(arc$\cot x \text{)} = - \frac{1}{1 + {x}^{2}}$
d/(dx)(arc$\sec x \text{)} = \frac{1}{| x | \sqrt{{x}^{2} - 1}}$
d/(dx)(arc$\csc x \text{)} = - \frac{1}{| x | \sqrt{{x}^{2} - 1}}$

Notice how $\cos x$, $\cot x$, and $\csc x$ are all negative, and otherwise the other trig function ($\sin x$, $\tan x$, $\sec x$) is the same.

That's all you need to do these.

a)
$\frac{d}{\mathrm{dx}} \left({\log}_{4} \left(6 x - 4\right)\right) = \frac{d}{\mathrm{dx}} \frac{\ln \left(6 x - 4\right)}{\ln 4} = \frac{6}{\left(6 x - 4\right) \ln 4}$

b)
You will need the quotient rule in here somewhere, along with the chain rule.

$\frac{d}{\mathrm{dx}} \left[{\log}_{3} \left({x}^{2} / \left(x - 4\right)\right)\right] = \frac{d}{\mathrm{dx}} \left[\frac{\ln \left({x}^{2} / \left(x - 4\right)\right)}{\ln 3}\right]$

$= \frac{1}{{x}^{2} / \cancel{x - 4}} \cdot \left[\frac{\left(x - 4\right) \left(2 x\right) - {x}^{2} \left(1\right)}{x - 4} ^ \cancel{2}\right] \cdot \frac{1}{\ln} 3$

Multiply the numerator out, subtract the ${x}^{2}$ terms, and factor out $x$:

$= \frac{1}{x} ^ \cancel{2} \cdot \left[\frac{\cancel{x} \left(x - 8\right)}{x - 4}\right] \cdot \frac{1}{\ln} 3$

$= \frac{1}{x} \cdot \left[\frac{x - 8}{x - 4}\right] \cdot \frac{1}{\ln} 3$

$= \frac{x - 8}{x - 4} \cdot \frac{1}{x \ln 3}$

c)
You can just use the product rule here, with what we've derived above. There are also some creative exponent rules here.

$\frac{d}{\mathrm{dx}} \left(x \cdot {4}^{- 2 x}\right) = x \cdot \left(- 2 \cdot {4}^{- 2 x} \ln 4\right) + {4}^{- 2 x} \cdot 1$

$= {4}^{- 2 x} \left[\left(x \cdot \left(- 2 \ln 4\right) + 1\right)\right]$

$= {4}^{- 2 x} \left(1 - 2 x \ln 4\right)$

or

${16}^{- x} \left(1 - 2 x \ln 4\right)$

d)

Reference the above, and use the chain rule.

$\frac{d}{\mathrm{dx}} \left(\arccos \left(4 \sqrt{x}\right)\right)$

$= - \frac{1}{\sqrt{1 - 16 x}} \cdot 4 \cdot \frac{1}{2 \sqrt{x}}$

$= - {\cancel{4}}^{2} / \left(\cancel{2} \sqrt{x} \sqrt{1 - 16 x}\right)$

$= - \frac{2}{\sqrt{x} \sqrt{1 - 16 x}}$

e)

Again, reference the above and use the chain rule.

$\frac{d}{\mathrm{dx}} \left[x \arctan \left(4 x - 6\right)\right] = x \cdot \left[\frac{1}{1 + {\left(4 x - 6\right)}^{2}}\right] \cdot 4 + \arctan \left(4 x - 6\right) \cdot 1$

$= \frac{4 x}{1 + {\left(4 x - 6\right)}^{2}} + \arctan \left(4 x - 6\right)$

or from Wolfram Alpha:
$= \frac{4 x}{{\left(6 - 4 x\right)}^{2} + 1} - \arctan \left(6 - 4 x\right)$

because $\arctan \left(u\right) = - \arctan \left(- u\right)$.