Question

Logarithmic differentiate cos^7(θ)

Answer 1

i know the general power rule of diffe' method yields

\[7 \cos ^6 (\theta). Dx(\cos(\theta) = -7 \sin(\theta)Cos^6(\theta)\]

which should be the same as the log diffe' method..i just need help on what the log diffe' would look like..i tend to mix it up

Answer 2

Let \(\large y=\cos^7\theta\). To compute the derivative via logarithmic differentiation, you first take natural logs of both sides to get \(\large \ln y = \ln (\cos^7\theta) \implies \ln y = 7\ln(\cos\theta)\).
Next you implicity differentiate both sides with respect to \(\large \theta\). Can you take things from here?

I hope this made sense! :-)

Answer 3

i know but do you mind finishing :/

Answer 4

Sure; implicitly differentiating both sides with respect to \(\large \theta\) gives us

\[\large \begin{aligned}\frac{d}{d\theta}\ln y = \frac{d}{d\theta} 7\ln(\cos\theta) &\implies \frac{1}{y}\frac{dy}{d\theta} = 7\frac{1}{\cos\theta}\cdot (-\sin\theta)\\ &\implies \frac{1}{y}\frac{dy}{d\theta} = -7\frac{\sin\theta}{\cos\theta}\\ &\implies\phantom{\frac{1}{y}}\frac{dy}{d\theta} = -7y\frac{\sin\theta}{\cos\theta} = -7\cos^6\theta\sin\theta\end{aligned}\]
since \(\large y=\cos^7\theta\).

Does this make sense? :-)

Answer 5

yup
thanks man