Science:Math Exam Resources/Courses/MATH100/December 2010/Question 01 (h)/Solution 1

To differentiate this, we should use logarithmic differentiation. First, we rewrite the function:

$\ln (f(x))=\ln ((\tan x{)}^{\cos x})=\cos (x)\ln (\tan x)$

And then use implicit differentiation

$\begin{array}{rl}(\ln (f(x)){)}^{\prime }& =(\cos (x)\ln (\tan (x)){)}^{\prime }\\ \frac{1}{f(x)}{f}^{\prime }(x)& =\cos (x)(\ln (\tan (x)){)}^{\prime }-\sin (x)\ln (\tan (x))\\ & =\cos (x)\frac{{\sec }^2(x)}{\tan (x)}-\sin (x)\ln (\tan (x))\end{array}$

(Remember that the derivative of $\tan (x)$ is ${\sec }^2(x)$). Now, simply solve for ƒ'(x) to obtain the answer:

$\begin{array}{rl}{f}^{\prime }(x)& =f(x)(\frac{\cos (x){\sec }^{}(x)}{\tan (x)}-\sin (x)\ln (\tan (x)))\\ & =(\tan (x){)}^{\cos (x)}(\frac{\sec (x)}{\tan (x)}-\sin (x)\ln (\tan (x)))\\ & =(\tan (x){)}^{\cos (x)}(\frac{1}{\sin (x)}-\sin (x)\ln (\tan (x)))\\ & =(\tan (x){)}^{\cos (x)}(\csc (x)-\sin (x)\ln (\tan (x)))\\ \end{array}$

(Note that $\cos (x){\sec }^{}(x)=\sec (x)$.)