Science:Math Exam Resources/Courses/MATH104/December 2014/Question 01 (k)

MATH104 December 2014

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Question 01 (k)
Find $f'(0)$, where $f(x)=\tan ^{-1}(5x).$ Note that $\tan ^{-1}$ refers to the inverse tangent function, which is also denoted by arctan.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
If you remember the derivative of $\tan ^{-1}{x}$ you can just use that directly. Don't forget the chain rule! Otherwise, take the tangent of both sides and then compare the derivatives. Again, don't forget the chain rule!

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution 1

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. The formula for derivative of $\tan ^{-1}(x)={\frac {1}{1+x^{2}}}$. Therefore, using the chain rule, we obtain that $[\tan ^{-1}(5x)]'={\frac {1}{1+(5x)^{2}}}(5)={\frac {5}{1+25x^{2}}}$. Plugging in $x=0$ we find that the final answer is $f'(0)={\frac {5}{1+25(0)^{2}}}=5$

Solution 2

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We can also solve this without using the derivative of $\tan ^{-1}x$ directly. Instead, let's taking the tangent of both sides yields $\tan {f(x)}=5x$ Comparing the derivatives - remember the chain rule - we see that ${\begin{aligned}(5x)'&=\left(\tan(f(x))\right)'\\5&=\left({\frac {\sin(f(x))}{\cos(f(x))}}\right)'\\&={\frac {\cos(f(x))\cos(f(x))f'(x)-\sin(f(x))(-\sin(f(x)))f'(x)}{\cos ^{2}(f(x))}}\\&={\frac {(\cos ^{2}(f(x))+\sin ^{2}(f(x)))f'(x)}{\cos ^{2}(f(x))}}\\&={\frac {f'(x)}{\cos ^{2}(f(x))}}\\5\cos ^{2}(f(x))&=f'(x)\\5\cos ^{2}(\tan ^{-1}(5x))&=f'(x)\end{aligned}}$ Finally, we plug in $x=0$ and use that $\tan ^{-1}(0)=0$ to obtain the final answer ${\begin{aligned}f'(0)&=5\cos ^{2}(\tan ^{-1}(0))\\&=5\cos ^{2}(0)\\&=5\end{aligned}}$

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