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:

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

And then use implicit differentiation

$\displaystyle {\begin{aligned}(\ln(f(x)))'&=(\cos(x)\ln {(\tan(x))})'\\{\frac {1}{f(x)}}f'(x)&=\cos(x)(\ln {(\tan(x))})'-\sin(x)\ln {(\tan(x))}\\&=\cos(x){\frac {\sec ^{2}(x)}{\tan(x)}}-\sin(x)\ln {(\tan(x))}\end{aligned}}$

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

${\begin{aligned}f'(x)&=f(x)\left({\frac {\cos(x)\sec ^{2}(x)}{\tan(x)}}-\sin(x)\ln {(\tan(x))}\right)\\&=(\tan(x))^{\cos(x)}\left({\frac {\sec(x)}{\tan(x)}}-\sin(x)\ln {(\tan(x))}\right)\\&=(\tan(x))^{\cos(x)}\left({\frac {1}{\sin(x)}}-\sin(x)\ln {(\tan(x))}\right)\\&=(\tan(x))^{\cos(x)}\left(\csc(x)-\sin(x)\ln {(\tan(x))}\right)\\\end{aligned}}$

(Note that $\cos(x)\sec ^{2}(x)=\sec(x)$.)