Tutorial 1: Basic Functions & Derivatives
Solving using MATLAB Solving using Python Solving using Mathematica
Q1: Find the limit of \(\lim _{x \rightarrow 5} \frac{x^{2}-25}{x^{2}+x-30}\).
Solution
\[\require{cancel} \begin{aligned} \lim _{x \rightarrow 5} \frac{x^{2}-25}{x^{2}+x-30} &= \lim _{x \rightarrow 5} \frac{\cancel{(x-5)}(x+5)}{\cancel{(x-5)}(x+6)} \\ &= \lim _{x \rightarrow 5} \frac{x+5}{x+6} \\ &= \frac{5+5}{5+6} \\ &= \mathbf{\frac{10}{11}} \end{aligned}\]Q2: Find the limit of \(\lim _{x \rightarrow 9} \frac{\sqrt{x}-3}{x-9}\).
Solution
\[\begin{aligned} \lim _{x \rightarrow 9} \frac{\sqrt{x}-3}{x-9} &= \lim _{x \rightarrow 9} \frac{\sqrt{x}-3}{x-9}\times\frac{\sqrt{x}+3}{\sqrt{x}+3} \\ &= \lim _{x \rightarrow 9} \frac{\cancel{x-9}}{\cancel{(x-9)}(\sqrt{x}+3)} \\ &= \lim _{x \rightarrow 9} \frac{1}{(\sqrt{x}+3)} \\ &= \frac{1}{\sqrt{9}+3} \\ &= \mathbf{\frac{1}{6}} \end{aligned}\]Q3: Find the limit of \(\lim _{x \rightarrow 0} \frac{\sqrt{2+x}-\sqrt{2}}{x}\).
Solution
\[\begin{aligned} \lim _{x \rightarrow 0} \frac{\sqrt{2+x}-\sqrt{2}}{x} &= \lim _{x \rightarrow 0} \frac{\sqrt{2+x}-\sqrt{2}}{x} \times \frac{\sqrt{2+x}+\sqrt{2}}{\sqrt{2+x}+\sqrt{2}} \\ &= \lim _{x \rightarrow 0} \frac{(\sqrt{2+x})^2-(\sqrt{2})^2}{x(\sqrt{2+x}+\sqrt{2})} \\ &= \lim _{x \rightarrow 0} \frac{2+x-2}{x(\sqrt{2+x}+\sqrt{2})} \\ &= \lim _{x \rightarrow 0} \frac{\cancel{x}}{\cancel{x}(\sqrt{2+x}+\sqrt{2})} \\ &= \frac{1}{\sqrt{2}+\sqrt{2}} \\ &= \frac{1}{2\sqrt{2}}\times\frac{\sqrt{2}}{\sqrt{2}} \\ &= \mathbf{\frac{\sqrt{2}}{4}} \end{aligned}\]Q4: Find the limit of \(\lim _{\theta \rightarrow \frac{\pi}{2}} \frac{\tan{\theta}}{\sec{\theta}}\).
Solution
\[\begin{aligned} \lim _{\theta \rightarrow \frac{\pi}{2}} \frac{\tan{\theta}}{\sec{\theta}} &= \lim _{\theta \rightarrow \frac{\pi}{2}} \frac{(\frac{\sin{\theta}}{\cancel{\cos{\theta}}})}{(\frac{1}{\cancel{\cos{\theta}}})} \\ &= \lim _{\theta \rightarrow \frac{\pi}{2}} \sin{\theta} \\ &= \sin{\frac{\pi}{2}} \\ &= \mathbf{1} \end{aligned}\]Q5: Find the limit of \(\lim _{\theta \rightarrow 0} \frac{\cos{\theta}-1}{\sin{\theta}}\).
Solution
\[\begin{aligned} \lim _{\theta \rightarrow 0} \frac{\cos{\theta}-1}{\sin{\theta}} &= \lim _{\theta \rightarrow 0} \frac{\cos{\theta}-1}{\sin{\theta}} \times \frac{\theta}{\theta} \\ &= \lim _{\theta \rightarrow 0} \frac{\cos{\theta}-1}{\theta} \times \frac{\theta}{\sin{\theta}} \\ &= \lim _{\theta \rightarrow 0} \frac{\cos{\theta}-1}{\theta} \times \lim _{\theta \rightarrow 0} \frac{\theta}{\sin{\theta}} \\ &= \lim _{\theta \rightarrow 0} \frac{\cos{\theta}-1}{\theta} \times \frac{1}{\lim _{\theta \rightarrow 0} \frac{\sin{\theta}}{\theta}} \\ &= 0\times1 \\ &= \mathbf{0} \end{aligned}\]Q6: If \(2x\leq g(x)\leq x^2-x+2\) for all \(x\), evaluate \(\lim_{x \rightarrow 1} g(x)\).
Solution
\[\lim_{x \rightarrow 1} 2x \leq \lim_{x \rightarrow 1} g(x) \leq \lim_{x \rightarrow 1} (x^2-x+2)\]Therefore,
\[2\leq\lim_{x \rightarrow 1} g(x) \leq 2\]Hence,
\[\lim_{x \rightarrow 1} g(x) = \mathbf{2}\]Q7: Solve \(y'\) if \(y=\sqrt{3x^2-2x+3}\).
Solution
Let \(u=3x^2-2x+3\), therefore \(\frac{du}{dx}=6x-2\).
Substitute \(y=\sqrt{u}=u^{\frac{1}{2}}\), therefore \(\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}\).
Putting back together,
\[\begin{aligned} \frac{dy}{dx} &=\frac{dy}{du}\cdot\frac{du}{dx} \\ &= \frac{1}{2}u^{-\frac{1}{2}} \cdot 6x-2 \\ &= \frac{1}{2}(3x^2-2x+3)^{-\frac{1}{2}} \cdot 6x-2 \\ &= \frac{6x-2}{2\sqrt{3x^2-2x+3}} \\ &= \boldsymbol{\frac{3x-1}{\sqrt{3x^2-2x+3}}} \end{aligned}\]Q8: Solve \(y'\) if \(y=5\sqrt[3]{x^2+\sqrt{x^3}}\).
Solution
\(y=5\sqrt[3]{x^2+\sqrt{x^3}}\) can be rewrite as \(y=5(x^2+x^{\frac{3}{2}})^{\frac{1}{3}}\).
Let \(u=x^2+x^{\frac{3}{2}}\), therefore \(\frac{du}{dx}=2x+\frac{3}{2}x^{\frac{1}{2}}\).
Substitute \(y=5u^{\frac{1}{3}}\), therefore \(\frac{dy}{du}=\frac{5}{3}u^{-\frac{2}{3}}\).
Putting back together,
\[\begin{aligned} \frac{dy}{dx} &=\frac{dy}{du}\cdot\frac{du}{dx} \\ &= \frac{5}{3}u^{-\frac{2}{3}}\cdot\left(2x+\frac{3}{2}x^{\frac{1}{2}}\right) \\ &= \boldsymbol{\frac{5}{3}\left(x^2+x^{\frac{3}{2}}\right)^{-\frac{2}{3}}\cdot\left(2x+\frac{3}{2}x^{\frac{1}{2}}\right)} \\ \end{aligned}\]Alternatively, the equation can be written as \(\frac{5\left(2x+\frac{3}{2}\sqrt{x}\right)}{3\left(x^2+\sqrt{x^3}\right)^{\frac{2}{3}}} = \frac{5\left(\frac{2x}{3}+\frac{1}{2}\sqrt{x}\right)}{\left(x^2+\sqrt{x^3}\right)^{\frac{2}{3}}} = \frac{5\left(\frac{2x}{3}+\frac{\sqrt{x^3}}{2x}\right)}{\left(x^2+\sqrt{x^3}\right)^{\frac{2}{3}}}\)
Q9: Solve \(y'\) if \(y=\ln{\cos{x^2}}\).
Solution
Let \(u=\cos{x^2}\), therefore \(\frac{du}{dx}=(-\sin{x^2})(2x)\).
Substitute \(y=\ln{u}\), therefore \(\frac{dy}{du}=\frac{1}{u}\).
Putting back together,
\[\begin{aligned} \frac{dy}{dx} &= \frac{dy}{du}\cdot\frac{du}{dx} \\ &= \frac{1}{u}\cdot(-\sin{x^2})(2x) \\ &= -\frac{2x\sin{x^2}}{\cos{x^2}} \\ &= \boldsymbol{-2x\tan{x^2}} \end{aligned}\]Q10: Differentiate \(y=\log{(4+\cos{x})}\).
Solution
Note that \(\frac{d}{dx} (\log_{b} (x))=\frac{1}{\ln (b)\cdot x} .\ \Longrightarrow \frac{d}{dx} (\log_{10} (x))=\frac{1}{\ln (10)\cdot x} =\frac{1}{x} \cdot \log e\). (\(\displaystyle \frac{1}{\ln 10} =\log e\))
\[\begin{aligned} \frac{d}{dx}(\log{4+\cos{x}}) &= \frac{1}{4+\cos{x}}\cdot\log{e} \cdot\frac{d}{dx} (4+\cos{x}) \\ &= \frac{1}{4+\cos{x}}\cdot\log{e} \cdot(-\sin{x}) \\ &= \boldsymbol{\frac{-(\log{e})(\sin{x})}{4+\cos{x}}} \end{aligned}\]Q11: Find \(y'\) for \(10e^{2xy}=e^{15y}+e^{13x}\).
Solution
\[\begin{aligned} 10e^{2xy}&=e^{15y}+e^{13x} \\ 10e^{2xy}(2x\cdot y'+2y)&=e^{15y}(15y')+e^{13x}(13) \\ 10e^{2xy}(2x\cdot y'+2y)&=15y'e^{15y}+13e^{13x} \\ (20xe^{2xy}-15e^{15y})y'&=13e^{13x}-20ye^{2xy} \\ \boldsymbol{y'}&=\boldsymbol{\frac{13e^{13x}-20ye^{2xy}}{20xe^{2xy}-15e^{15y}}} \end{aligned}\]Q12: Find \(f'(x)\) if \(f(x)=2x(\arctan{5x})^2+6\tan{(\cos{6x})}\).
Solution
- \[\frac{d}{dx} (\tan^{-1}{5x})^2=2\tan^{-1}{5x}\cdot\frac{1}{1+(5x)^2}\cdot5=\frac{10\tan^{-1}{5x}}{1+25x^2}\]
- \[\frac{d}{dx} 2x(\tan^{-1}{5x})^2=2x(\frac{10\tan^{-1}{5x}}{1+25x^2})+(\tan^{-1}{5x})^2\cdot2=\frac{20x\tan^{-1}{5x}}{1+25x^2}+2(\tan^{-1}{5x})^2\]
- \[\frac{d}{dx} 6\tan{(\cos{6x})}=6\sec^2{(\cos{6x})} \frac{d}{dx}\cos{6x}=6\sec^2{(\cos{6x})}\cdot(-\sin{6x})\cdot6=-36(\sec^2{(\cos{6x})})\sin{6x}\]
Q13: Solve \(y'\) if \(y=4x\sinh^{-1}{(\frac{x}{6})}+\tanh^{-1}{(\cos{10x})}\).
Solution
- \[\frac{d}{dx} 4x\sinh^{-1}{(\frac{x}{6})}=4x\left[\frac{1}{\sqrt{1+(\frac{x}{6})^2}}\cdot\frac{1}{6}\right]+\sinh^{-1}{(\frac{x}{6})}\cdot4=\frac{\frac{2x}{3}}{\sqrt{1+\frac{x^2}{36}}}+4\sinh^{-1}{(\frac{x}{6})}\]
- \[\frac{d}{dx} \tanh^{-1}{(\cos{10x})}=\frac{1}{1-(\cos{10x})^2}\cdot\frac{d}{dx}\cos{10x}=\frac{1}{1-\cos^2{10x}}\cdot(-10\sin{10x})=\frac{-10\sin{10x}}{\sin^2{10x}}=-10\csc{10x}\]