# Thread: Differentiating using log question

1. ## Differentiating using log question

find ds/dt for s = cos^4t (1+t^2)^0.5

(^4 means to the power 4, so cos to power 4, and there's also root (1 plus t squared)

Problem is that I take logs of both sides then try to differentiate using the 1/s ds/dt = rule but get the wrong answer, so a worked answer would be much appreciated. Thanks for any help.

2. Originally Posted by wardj
find ds/dt for s = cos^4t (1+t^2)^0.5

(^4 means to the power 4, so cos to power 4, and there's also root (1 plus t squared)

Problem is that I take logs of both sides then try to differentiate using the 1/s ds/dt = rule but get the wrong answer, so a worked answer would be much appreciated. Thanks for any help.
$s=\cos^4(t)\sqrt{1+t^2}$

$\ln(s)=\ln\left(\cos^4(t)\sqrt{1+t^2}\right)$

$\ln(s)=4\ln\left(\cos(t)\right)+\frac{1}{2}\ln\lef t(1+t^2\right)$

Differentiating, we get:

$\frac{1}{s}\cdot\frac{ds}{dt}=\frac{4}{\cos(t)}\cd ot(-\sin(t))+\frac{1}{2(1+t^2)}\cdot(2t)$

$\frac{1}{s}\frac{ds}{dt}=-4\tan(t)+\frac{t}{1+t^2}$

$\frac{ds}{dt}=s\cdot\left[-4\tan(t)+\frac{t}{1+t^2}\right]$

Since $s=\cos^4(t)\sqrt{1+t^2}$,

$\color{red}\boxed{\frac{ds}{dt}=\cos^4(t)\sqrt{1+t ^2}\left[-4\tan(t)+\frac{t}{1+t^2}\right]}$

3. Originally Posted by wardj
find ds/dt for s = cos^4t (1+t^2)^0.5

(^4 means to the power 4, so cos to power 4, and there's also root (1 plus t squared)

Problem is that I take logs of both sides then try to differentiate using the 1/s ds/dt = rule but get the wrong answer, so a worked answer would be much appreciated. Thanks for any help.
$\log s = 4\log \cos t + \frac12 \log (1+t^2)$

Differentiate:
$\frac{s'}{s} = -4\frac{\sin t}{\cos t} + \frac12 \frac{2t}{1+t^2}$

$\frac{s'}{s} = -4\tan t + \frac{t}{1+t^2}$

$\frac{ds}{dt} = -4\sin t \cos^3 t \sqrt{1+t^2} + \frac{t\cos^3 t \sqrt{1+t^2}}{1+t^2}$

$\frac{ds}{dt} = -4 \cos^3 t \sin t \sqrt{1+t^2} + \frac{t\cos^3 t }{\sqrt{1+t^2}}$