Logarithmic Differentiation for Trig Functions?

Question

Can someone explain to me how logarithmic differentiation for the trig functions (Sin, Cosine, Tangent, Secant, Cosecant, Cotangent) are derived?

For instance, one of these problems that involves trig functions is:

(Note: I am using the "@" symbol in place of theta, so the problem reads as y = theta + 5/ theta*cos*theta)

y = @ + 5/ @cos@

I know the first step is:

lny = ln(@+5) -ln@ - ln(cos@)

What I don't get is how they are saying (in my student solution's manual) that the natural log of -ln(cos@) is + sin@/cos@, which the next step shows:

1/y dy/d@ = 1/(@ + 5) - 1/@ + sin@/cos@

I don’t get how they are getting this, as I thought it would simply be -1/cos@, or –sec@.

Can someone explain to me why this is the case for the cosine, and how the natural logs of the other five trig functions are derived and what they are as well? It would be very helpful.

Answer 1

Always use parentheses - it avoids confusion (as there was from one respondent).
We have:
y = (θ+5)/(θcosθ)

You always use logarithmic differentiation with a product or quotient: if y = a(x)b(x)/(c(x)d(x))
ln(y) = ln(a) + ln(b) - ln(c) - ln(d)
so
y'/y = a'/a + b'/b - c'/c -d'/d
or
y' = y(a'/a + b'/b - c'/c - d'/d)
In other words, term on the top have a plus sign, terms on the bottom have a minus sign.

On to your example (notice that you have one term on top (θ+5) and two on the bottom (θ and also cos(θ)):
y = (θ+5)/(θcosθ)
so
y' = y (1/(θ+5) - 1/θ - (-sin(θ)/cos(θ)))
= y (1/(θ+5) - 1/θ + sin(θ)/cos(θ))
= y (1/(θ+5) - 1/θ + tan(θ))

To directly address your question:
write down the derivatives of all the trig functions:
sin'(θ) = cos(θ)
cos'(θ) = -sin(θ)
tan'(θ) = sec²(θ)
sec'(θ) = tan(θ)sec(θ)
cot'(θ) = csc²(θ)
csc'(θ) = -cot(θ)csc(θ)

so
ln(sin(θ))' = cos(θ)/sin(θ) = cot(θ)
ln(cos(θ))' = -sin(θ)/cos(θ) = -tan(θ)
ln(tan(θ))' = sec²(θ)/tan(θ) = sin(θ)/cos³(θ)
ln(sec(θ))' = tan(θ)sec(θ)/sec(θ) = tan(θ)
ln(cot(θ))' = csc²(θ)/cot(θ) = cos(θ)/sin³(θ)
ln(csc(θ))' = -cot(θ)csc(θ)/csc(θ) = -cot(θ)

Answer 2

The derivative of the -ln(cos@) with respect to @ is in fact sin@/cos@.

The derivative of ln(u) is just du divided by u. In your case, u = cos@, so the derivative w.r.t. @ of ln(cos@) is -sin@/cos@. Combine that minus with the minus in front of the original expression and you get +sin@/cos@.

The confusion may be due to the fact that the derivative of ln(x) is 1/x. Well, that's just because the derivative of x is 1. In general, the numerator of the derivative of a log will be the derivative of whatever was inside the parenthesis following the log.

Answer 3

y=@+/cos@
lny=ln(@+5)-ln(cos@)
when we differentiat lny then we can write
dy/d@ = 1/y .dy/d@ (differentiating w.r.t @)
because the differentiation of any variable associated with ln should goto denominator after differentiation(proving which is not our task and beyond this level).
thats why the third step becomes as .i would like to suggest you to learn first the derivatives.
(any way this is solved with minset that the reader is already familiar with derivatives.)
since there is also an ln on the term @ and cos @
thus the step ln@-lncos@ becomes after differentiation as
1/@+5-1/@+sin@cos@
this is step you will easily understand and will feel relieved when you study derivatives any way i try a bit
the derivative of the lncos@ is
1/@+5-1/sin@cos@
the derivative of sin@ is dy/d@=dsin@/d@=cos@
the derivative of cos@ is dy/d@= dcos@/d@= -sin@
then we have solve the derivative furhter after the variable not given in the deivative like we have @
then the third step becomes as
1/y.dy/d@=1/(@+5) -1/(-sin@cos@)
=1/(@+5)+1/(sin@cos@)

Answer 4

I shall use A for the angle as it doesn't matter what you use.
-log(cosA) has to be differentiated by the chain rule.
Let z = cosA so dz/dA = -sinA
if r = -logz then dr/dz = -1/z = -1/cosA
so dr/dz = dr/dz*dz/dA = (-1/cosA)*(-sinA) = +sinA/cosA

If you do want to know how the ln's of other trig differentiates then they are
ln(sinA) becomes cosA/sinA
ln(tanA) becomes (secA)^2/tanA
ln(cotA) becomes -(cscA)^2/cotA
ln(secA) becomes secAtanA/secA = tanA
ln(cscA) becomes -cscAcotA/cscA = -cotA