logarithmic differentiation

Question

If p(x)=(2x+4)^tanx, find p′(x).  Enter an exact answer in terms of a natural logarithm.

Answer 1

I'll use "y" in place of p(x) and change it back at the end:

y = (2x + 4)^tan(x)

ln(y) = ln[(2x + 4)^tan(x)]

ln(y) = tan(x)•ln(2x + 4))

Now differentiate. Remember that we are differentiating with respect to "x" so the differentiate the left side, ln(y), we must use the chain rule.

[ln(y)] ' = (1/y)•y'

The right side is a product of "tan(x)" and "ln(2x + 4) so we must use the product rule (u•v)' = u'v + uv'

So [tan(x)•ln(2x + 4))] ' = (tan(x))' • ln(2x + 4) + tan(x)•[ln(2x + 4)] '

(tan(x))' = sec²(x) and [ln(2x + 4)] ' = 1/(2x + 4) • 2 = 2/(2x + 4) = 1/(x + 2) so putting those together:

[tan(x)•ln(2x + 4))] ' = sec²(x) • ln(2x + 4) + tan(x)•(1/(x + 2)) = sec²(x) • ln(2x + 4) + tan(x)/(x + 2)

So:

(1/y)•y' = sec²(x) • ln(2x + 4) + tan(x)/(x + 2)

y' = y[sec²(x) • ln(2x + 4) + tan(x)/(x + 2)]

p'(x) = (2x + 4)^tan(x)[sec²(x) • ln(2x + 4) + tan(x)/(x + 2)]