Science:Math Exam Resources/Courses/MATH100/December 2012/Question 02 (c)/Solution 2

For an alternative solution, taking the sin of both sides yields

$\displaystyle \sin(y)=\ln(x)$

Taking implicit derivatives of both sides yields

$\displaystyle \cos(y)y'={\frac {1}{x}}$

Solving for the derivative yields

$\displaystyle y'={\frac {1}{\cos(y)x}}$

Next, we want to express the right hand side as a function of x. Since $\displaystyle \sin(y)=\ln(x)$ and $\displaystyle \sin(y)$ is opposite over hypotenuse, we can draw the following triangle, interpreting y as an angle:

where the last side (the base of the triangle) is computed using the Pythagorean theorem. Thus we see that

$\displaystyle \cos(y)={\sqrt {1-(\ln(x))^{2}}}$

and plugging into the derivative yields

$\displaystyle y'={\frac {1}{x{\sqrt {1-(\ln(x))^{2}}}}}$

completing the problem.