Answers to Calculus Practical 5

Note: below is a list of corrections to the solutions provided by the lecturer, but everything I have written here is correct, I think.

• Question 5a: error in answer, should be $3a{x}^2+b$ not $a{x}^2+b$
• Question 6: easier route through logarithmic differentiation

Question 1 §

Question 1a §

$$f(x)=3x^5-2x^2+7x^{-\frac{1}{2}}+2⟹f^{\prime }(x)=15x^4-4x-\frac{7}{2}{x}^{-\frac{3}{2}}$$

Question 1b §

$$f(x)=\frac{x^3+3x^2+7}{x-1}⟹f^{\prime }(x)=\frac{(3x^2+6x)(x-1)-(1)(x^3+3x^2+7)}{(x-1{)}^2}$$

$$⟹f^{\prime }(x)=\frac{2x^3-6x-7}{(x-1{)}^2}$$

Question 1c §

$$f(x)=e^x\sin x⟹f^{\prime }(x)=e^x\sin x+e^x\cos x$$

$$⟹f^{\prime }(x)=e^x(\sin x+\cos x)$$

Question 1d §

$$f(x)=4x^4{e}^{cx}⟹f^{\prime }(x)=16x^3{e}^{cx}+4c{x}^4{e}^{cx}$$

$$⟹f^{\prime }(x)=4x^3{e}^{cx}(4+x)$$

Question 1e §

$$f(x)=\frac{\cos ax}{x^3}⟹f^{\prime }(x)=-\frac{a{x}^3\sin ax-3x^2\cos ax}{x^6}$$

$$⟹f^{\prime }(x)=\frac{3\cos ax-ax\sin ax}{x^4}$$

Question 1f §

$$f(x)=\frac{a{e}^{bx}}{\sin cx}⟹f^{\prime }(x)=\frac{ab{e}^{bx}\sin cx-ac{e}^{bx}\cos cx}{{\sin }^{2}cx}$$

$$⟹f^{\prime }(x)=\frac{a{e}^{bx}(b\sin cx-c\cos cx)}{{\sin }^{2}cx}$$

Question 2 §

Question 2a §

$$f(x)=\sin (5x+2)⟹f^{\prime }(x)=5\cos (5x+2)$$

Question 2b §

$$f(x)=\ln \left(1+x^{-2}\right)⟹f^{\prime }(x)=-\frac{2x^{-3}}{1+x^{-2}}$$

$$⟹f^{\prime }(x)=-\frac{2}{x(x^2+1)}$$

Question 2c §

$$f(x)={\sin }^{3}x⟹f^{\prime }(x)=3\cos x{\sin }^{2}x$$

Question 3 §

Question 3a §

$$h(t)=(t^3-1{)}^{100}⟹h^{\prime }(t)=300t^2(t^3-1{)}^{99}$$

Question 3b §

$$h(t)=\sin (a+b{t}^4)⟹h^{\prime }(t)=4b{t}^3\cos (a+b{t}^4)$$

Question 3c §

$$h(t)=a\cos (b\tan ct)⟹h^{\prime }(t)=-abc{\sec }^2(ct)\sin (b\tan ct)$$

Question 4 §

Question 4a §

$$\frac{d}{dx}\left(x^4{e}^{3x}\tan x\right)⟹\frac{d}{dx}\left(4\ln x+3x+\ln \tan x\right)$$

$$⟹x^4{e}^{3x}\tan x\cdot (\frac{4}{x}+3+2\csc (2x))$$

Question 4b §

$$\frac{d}{dx}\left(\frac{e^{4x}}{x^3\cosh 2x}\right)⟹\frac{d}{dx}(4x-3\ln x-\ln \cosh 2x)$$

$$⟹\frac{e^{4x}}{x^3\cosh 2x}\cdot (4-\frac{3}{x}-\tanh 2x)$$

Question 5 §

Question 5a §

$$f(x)=a{x}^3+bx:f^{\prime }(x)=\underset{h\to 0}{\lim }\left(\frac{f(x+h)-f(x)}{h}\right)$$

$$⟹f^{\prime }(x)=\underset{h\to 0}{\lim }\left(\frac{(a(x+h{)}^3+b(x+h))-(a{x}^3+bx)}{h}\right)$$

$$⟹f^{\prime }(x)=\underset{h\to 0}{\lim }\left(\frac{a((x+h{)}^3-x^3)+bh}{h}\right)$$

$$⟹f^{\prime }(x)=\underset{h\to 0}{\lim }\left(\frac{bh+a((x+h)-x)((x+h{)}^2+x(x+h)+x^2)}{h}\right)$$

$$⟹f^{\prime }(x)=\underset{h\to 0}{\lim }\left(b+a(3x^2+h^2+3xh)\right)$$

$$⟹f^{\prime }(x)=3a{x}^2+b$$

Question 5b §

$$f(x)=\sqrt{cx+d}:f^{\prime }(x)=\underset{h\to 0}{\lim }\left(\frac{f(x+h)-f(x)}{h}\right)$$

$$⟹f^{\prime }(x)=\underset{h\to 0}{\lim }\left(\frac{(\sqrt{c(x+h)+d})-(\sqrt{cx+d})}{h}\right)$$

$$⟹f^{\prime }(x)=\underset{h\to 0}{\lim }\left[\left(\frac{\sqrt{(cx+d)+ch}-\sqrt{cx+d}}{h}\right)\left(\frac{\sqrt{(cx+d)+ch}+\sqrt{cx+d}}{\sqrt{(cx+d)+ch}+\sqrt{cx+d}}\right)\right]$$

$$⟹f^{\prime }(x)=\underset{h\to 0}{\lim }\left(\frac{c}{\sqrt{cx+d+ch}+\sqrt{cx+d}}\right)$$

$$\therefore f^{\prime }(x)=\frac{c}{2\sqrt{cx+d}}$$

Question 5c (to-do) §

$$f(x)=\frac{1}{\sqrt{ex+f}}:f^{\prime }(x)=\underset{h\to 0}{\lim }\left(\frac{f(x+h)-f(x)}{h}\right)$$

$$⟹f^{\prime }(x)=\underset{h\to 0}{\lim }\left(\frac{(e(x+h)+f{)}^{-\frac{1}{2}}-(ex+f{)}^{-\frac{1}{2}}}{h}\right)$$

$$⟹f^{\prime }(x)=\underset{h\to 0}{\lim }\left(\frac{}{h}\right)$$

Solution: Pasted image 20231027195115.png

Question 6 §

$$f(x)=(x-a)(x-b)(x-c)⟹f(x)=\ln (x-a)+\ln (x-b)+\ln (x-c)$$

$$⟹f^{\prime }(x)=(x-a)(x-b)(x-c)\cdot \left(\frac{1}{x-a}+\frac{1}{x-b}+\frac{1}{x-c}\right)$$

$$\therefore \frac{f^{\prime }(x)}{f(x)}=\frac{1}{x-a}+\frac{1}{x-b}+\frac{1}{x-c}\square$$

Question 7 §

Question 7a §

$$\cos 2x={\cos }^{2}x-{\sin }^{2}x⟹-2\sin 2x=-2\sin x\cos x-2\sin x\cos x$$

$$⟹\sin 2x=2\sin x\cos x$$

Question 7b §

$$\sin (a+b)=\sin a\cos b+\cos a\sin b$$

$$⟹\left(1+\frac{db}{da}\right)\cos (a+b)=\left(\cos a\cos b-\frac{db}{da}\sin a\sin b\right)+\left(-\sin a\sin b+\frac{db}{da}\cos a\cos b\right)$$

$$⟹\cos (a+b)=\frac{\left(1+\frac{da}{db}\right)(\cos a\cos b-\sin a\sin b)}{1+\frac{db}{da}}$$

$$\therefore \cos (a+b)=\cos a\cos b-\sin a\sin b$$