Skip to content
Exercise 8.2 · Q2

Q.Find dfdf for f(x)=x2+3xf(x)=x^2+3x and evaluate it for

(i) x=2x=2 and dx=0.1dx=0.1
(ii) x=3x=3 and dx=0.02dx=0.02
Tamil Nadu DgeTextbookSubjectiveImportance★★★★★
9% · 9/99 Questions
✓ Free question

f(x)=x2+3x⇒f′(x)=2x+3f(x)=x^2+3x\Rightarrow f'(x)=2x+3, so df=(2x+3) dxdf=(2x+3)\,dx; substitute each given (x,dx)(x,dx) pair.

Step 1. General differential. f′(x)=2x+3f'(x)=2x+3, so df=(2x+3) dxdf=(2x+3)\,dx.

Step 2. Part (i): x=2, dx=0.1x=2,\ dx=0.1. df=(2(2)+3)(0.1)=(4+3)(0.1)=7(0.1)=0.7df=(2(2)+3)(0.1)=(4+3)(0.1)=7(0.1)=0.7.

Step 3. Part (ii): x=3, dx=0.02x=3,\ dx=0.02. df=(2(3)+3)(0.02)=(6+3)(0.02)=9(0.02)=0.18df=(2(3)+3)(0.02)=(6+3)(0.02)=9(0.02)=0.18.

✓Final answer

(i) df=0.7df=\boxed{0.7} (ii) df=0.18df=\boxed{0.18}

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.