Question

Find the instantaneous rate of change of the function f(x) = x2 + 3x + 4 when x = 2 using First Principles. Confirm that your answer is correct using the derivative rules.

Answer 1

$Solution: ~ \\Given: f(x)=x^2 +3x+4 \\To ~Find :~derivative~At~ x=2~ by~ first ~ principle~and~ confirm ~ using~ \\derivative ~rules. \\ Solution: ~Since ~f(x)=x^2+3x+4 \\ \therefore f(x+h)=(x+h)^2 +3(x+h)+4 \\~~~~~~~~~~~~~~~~~~~~=x^2+2xh+h^2+3x+3h+4 \\Now~By~ first ~ principle~ we ~ know~that, \\f'(x)=\lim_{h \to 0} f(x) = \frac{f(x+h)-f(x)}{h} \\ \therefore f'(x)=\lim_{h \to 0} \frac{(x^2+2xh+h^2+3x+3h+4)-(x^2+3x+4)}{h} \\~~~~~~~~~~~~~~=\lim_{h \to 0} \frac{(x^2+2xh+h^2+3x+3h+4-x^2-3x-4)}{h} \\~~~~~~~~~~~~~~=\lim_{h \to 0} \frac{2xh+h^2+3h}{h} \\~~~~~~~~~~~~~~=\lim_{h \to 0} \frac{h(2x+h+3)}{h} \\~~~~~~~~~~~~~~=\lim_{h \to 0} 2x+h+3 \\~~~~~~~~~~~~~~=2x+0+3 \\~~~~~~~~~~~~~~=2x+3 \\\therefore f'(x)=2x+3 \\at ~x=2, \\f'(2)=2(2)+3=4+3=7 \\ \therefore f'(x)=7 \\By~using~derivative~rule~, \\we~know~that~f'(x)=\frac{d}{dx}f(x)=\frac{d}{dx}(x^n)=n.x^{n-1}~ and ~\frac{d}{dx}f(x)=\frac{d}{dx}(k)=0 \\ \therefore f'(x)=\frac{d}{dx}f(x)=\frac{d}{dx}(x^2 +3x+4)=\frac{d}{dx}(x^2) +\frac{d}{dx}(3x)+\frac{d}{dx}(4) \\f'(x)=2x^{2-1}+3^{1-1}+0=2x+3 \\Now, at ~x=2, \\f'(2)=2(2)+3=4+3=7 \\ \therefore f'(x)=7 \\Hence~ the~ instantaneous ~rate~ of~ change~ of~ the ~ function~ f(x)~ at ~x=2 ~is~ 7.$