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Exercise 10.1 · Q1

Q.Find the derivatives of the following functions using first principle.

(i) f(x)=6f(x) = 6
(ii) f(x)=−4x+7f(x) = -4x + 7
(iii) f(x)=−x2+2f(x) = -x^2 + 2
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Step 1. Write the first-principle definition: f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x)=\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}.

Step 2. (i) f(x)=6f(x)=6. Then f(x+h)=6f(x+h)=6 (a constant function), so f(x+h)−f(x)=6−6=0f(x+h)-f(x)=6-6=0. Hence f(x+h)−f(x)h=0h=0\dfrac{f(x+h)-f(x)}{h}=\dfrac{0}{h}=0 for every h≠0h\neq0, so f′(x)=lim⁡h→00=0f'(x)=\lim_{h\to0}0=0.

Step 3. (ii) f(x)=−4x+7f(x)=-4x+7. Then f(x+h)=−4(x+h)+7=−4x−4h+7f(x+h)=-4(x+h)+7=-4x-4h+7. So f(x+h)−f(x)=(−4x−4h+7)−(−4x+7)=−4hf(x+h)-f(x)=(-4x-4h+7)-(-4x+7)=-4h. Dividing by hh: −4hh=−4\dfrac{-4h}{h}=-4. Hence f′(x)=lim⁡h→0(−4)=−4f'(x)=\lim_{h\to0}(-4)=-4.

Step 4. (iii) f(x)=−x2+2f(x)=-x^2+2. Then f(x+h)=−(x+h)2+2=−x2−2xh−h2+2f(x+h)=-(x+h)^2+2=-x^2-2xh-h^2+2. So f(x+h)−f(x)=(−x2−2xh−h2+2)−(−x2+2)=−2xh−h2=h(−2x−h)f(x+h)-f(x)=(-x^2-2xh-h^2+2)-(-x^2+2)=-2xh-h^2=h(-2x-h). Dividing by hh: −2x−h-2x-h. Hence f′(x)=lim⁡h→0(−2x−h)=−2xf'(x)=\lim_{h\to0}(-2x-h)=-2x.

✓Final answer

(i) f′(x)=0f'(x)=0 (ii) f′(x)=−4f'(x)=-4 (iii) f′(x)=−2xf'(x)=-2x

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