Question

1) Differentiate the following function
g(x) = e^-5x + ((1/x)- tanx)^3 ( squat root of x^2 +1)

2)
Let f(x)= pi- squat root of 7x-9. Use the definition of the derivative to find f’(x)

Answer 1

Solution: 1) Given that $g(x)= e^{-5x}+(\frac{1}{x}-tan x^3) (\sqrt{x^2+1})$

$\therefore g'(x)=\frac{d}{dx}[ e^{-5x}+(\frac{1}{x}-tan x^3) (\sqrt{x^2+1})]$

$g'(x)=\frac{d}{dx}[ e^{-5x}]+\frac{d}{dx}[(\frac{1}{x}-tan x^3) (\sqrt{x^2+1})]$

$g'(x)=e^{-5x}\frac{d}{dx}[ -5x]+(\sqrt{x^2+1})\frac{d}{dx}[(\frac{1}{x}-tan x)^3]+[(\frac{1}{x}-tan x)^3]\frac{d}{dx}[\sqrt{x^2+1}]$

$g'(x)=e^{-5x}[-5\frac{d}{dx}(x)+(\sqrt{x^2+1}).3(\frac{1}{x}- tan(x))^2\frac{d}{dx}[(\frac{1}{x}-tan(x))]+[(\frac{1}{x}-tan (x))^3].\frac{1}{2}(x^2+1)^{\frac{1}{2}-1}.\frac{d}{dx}[x^2+1]$

$g'(x)=-5e^{-5x}+(\sqrt{x^2+1}).3(\frac{1}{x}- tan(x))^2(\frac{d}{dx}[\frac{1}{x}]-\frac{d}{dx}[tan(x)])+[(\frac{1}{x}-tan(x))^3].\frac{1}{2}(x^2+1)^{-\frac{1}{2}}.(\frac{d}{dx}[x^2]+\frac{d}{dx}[1])$

$g'(x)=-5e^{-5x}+(\sqrt{x^2+1}).3(\frac{1}{x}- tan(x))^2(-\frac{1}{x^2}-sec^2 x)+[(\frac{1}{x}-tan (x))^3].\frac{1}{2\sqrt{x^2+1}}(2x+0)$

$g'(x)=-5e^{-5x}+(\sqrt{x^2+1}).3(\frac{1}{x}- tan(x))^2(-\frac{1}{x^2}-sec^2 x)+.\frac{(2x)[(\frac{1}{x}-tan (x))^3]}{2\sqrt{x^2+1}}$

$g'(x)=-5e^{-5x}+(\sqrt{x^2+1}).3(\frac{1}{x}- tan (x))^2(-\frac{1}{x^2}-sec^2 x)+.\frac{(x)(\frac{1}{x}-tan (x))^3}{\sqrt{x^2+1}}$

Solution: 2) Given that $f(x)=\pi-\sqrt{7x-9}$

By definition of derivative, $f'(x)=\frac{f(x+h)-f(x)}{h}$

$\therefore f'(x)= lim_{h\to0}\frac{\pi-\sqrt{7(x+h)-9} - (\pi-\sqrt{7x-9})}{h}$

$f'(x)= lim_{h\to0}\frac{\pi-\sqrt{7(x+h)-9} - \pi+\sqrt{7x-9})}{h}$ [brackets opening]

$f'(x)= lim_{h\to0}\frac{-\sqrt{7(x+h)-9} +\sqrt{7x-9})}{h}$

$f'(x)= lim_{h\to0}\frac{\sqrt{7x-9}-\sqrt{7(x+h)-9}}{h}$

$f'(x)= lim_{h\to0}\frac{\sqrt{7x-9}-\sqrt{7(x+h)-9}}{h}.\frac{\sqrt{7x-9}+\sqrt{7(x+h)-9}}{\sqrt{7x-9}+\sqrt{7(x+h)-9}}$ [Rationalization]

$f'(x)= lim_{h\to0}\frac{(\sqrt{7x-9}-\sqrt{7(x+h)-9})(\sqrt{7x-9}+\sqrt{7(x+h)-9})}{h(\sqrt{7x-9}+\sqrt{7(x+h)-9})}$

$f'(x)= lim_{h\to0}\frac{(\sqrt{7x-9})^2+(\sqrt{7(x+h)-9})^2}{h(\sqrt{7x-9}+\sqrt{7(x+h)-9})}$ [using formula (a-b)(a+b)=a²-b²]

$f'(x)= lim_{h\to0}\frac{7x-9-7x-7h+9}{h(\sqrt{7x-9}+\sqrt{7(x+h)-9})}$

$f'(x)= lim_{h\to0}\frac{-7h}{h(\sqrt{7x-9}+\sqrt{7(x+h)-9})}$

$f'(x)= lim_{h\to0}\frac{-7}{(\sqrt{7x-9}+\sqrt{7(x+h)-9})}$

$f'(x)=\frac{-7}{(\sqrt{7x-9}+\sqrt{7x-9})}$

$f'(x)=\frac{-7}{2\sqrt{7x-9}}$