Derivatives #17 - Questions and Answers

Question 1 Find the derivative of 𝑓 ( 𝑥 ) = 𝑥 cos ⁡ 𝑥 1 + 𝑥 2 f(x)= 1+x 2 ​ xcosx ​ : Solution: Use the quotient rule: If 𝑓 ( 𝑥 ) = 𝑢 𝑣 f(x)= v u ​ , then: 𝑓 ′ ( 𝑥 ) = 𝑣 𝑑 𝑢 𝑑 𝑥 − 𝑢 𝑑 𝑣 𝑑 𝑥 𝑣 2 . f ′ (x)= v 2 v dx du ​ −u dx dv ​ ​ . Identify 𝑢 u and 𝑣 v: 𝑢 = 𝑥 cos ⁡ 𝑥 u=xcosx, so 𝑑 𝑢 𝑑 𝑥 = cos ⁡ 𝑥 − 𝑥 sin ⁡ 𝑥 dx du ​ =cosx−xsinx (product rule). 𝑣 = 1 + 𝑥 2 v= 1+x 2 ​ , so 𝑑 𝑣 𝑑 𝑥 = 𝑥 1 + 𝑥 2 dx dv ​ = 1+x 2 ​ x ​ (chain rule). Apply the quotient rule: 𝑓 ′ ( 𝑥 ) = 1 + 𝑥 2 ⋅ 𝑑 𝑑 𝑥 ( 𝑥 cos ⁡ 𝑥 ) − ( 𝑥 cos ⁡ 𝑥 ) ⋅ 𝑑 𝑑 𝑥 ( 1 + 𝑥 2 ) ( 1 + 𝑥 2 ) 2 . f ′ (x)= ( 1+x 2 ​ ) 2 1+x 2 ​ ⋅ dx d ​ (xcosx)−(xcosx)⋅ dx d ​ ( 1+x 2 ​ ) ​ . Substitute 𝑑 𝑢 𝑑 𝑥 dx du ​ and 𝑑 𝑣 𝑑 𝑥 dx dv ​ : 𝑓 ′ ( 𝑥 ) = 1 + 𝑥 2 ( cos ⁡ 𝑥 − 𝑥 sin ⁡ 𝑥 ) − 𝑥 cos ⁡ 𝑥 ( 𝑥 1 + 𝑥 2 ) 1 + 𝑥 2 . f ′ (x)= 1+x 2 1+x 2 ​ (cosx−xsinx)−xcosx( 1+x 2 ​ x ​ ) ​ . Simplify the terms step by step to get: 𝑓 ′ ( 𝑥 ) = cos ⁡ 𝑥 ( 1 + 𝑥 2 ) − 𝑥 sin ⁡ 𝑥 ( 1 + 𝑥 2 ) − 𝑥 2 cos ⁡ 𝑥 ( 1 + 𝑥 2 ) 3 / 2 . f ′ (x)= (1+x 2 ) 3/2 cosx(1+x 2 )−xsinx(1+x 2 )−x 2 cosx ​ . Question 2 Solve the derivative 𝑥 𝑦 = 𝑒 2 𝑥 xy=e 2x : Solution: Differentiate both sides with respect to 𝑥 x: Use the product rule for 𝑥 𝑦 xy: 𝑑 𝑑 𝑥 ( 𝑥 𝑦 ) = 𝑥 𝑑 𝑦 𝑑 𝑥 + 𝑦 dx d ​ (xy)=x dx dy ​ +y. Differentiate 𝑒 2 𝑥 e 2x : 𝑑 𝑑 𝑥 ( 𝑒 2 𝑥 ) = 2 𝑒 2 𝑥 dx d ​ (e 2x )=2e 2x . 𝑥 𝑑 𝑦 𝑑 𝑥 + 𝑦 = 2 𝑒 2 𝑥 . x dx dy ​ +y=2e 2x . Solve for 𝑑 𝑦 𝑑 𝑥 dx dy ​ : 𝑥 𝑑 𝑦 𝑑 𝑥 = 2 𝑒 2 𝑥 − 𝑦 . x dx dy ​ =2e 2x −y. 𝑑 𝑦 𝑑 𝑥 = 2 𝑒 2 𝑥 − 𝑦 𝑥 . dx dy ​ = x 2e 2x −y ​ . Question 3 Prove 𝑑 𝑑 𝑥 ( cot ⁡ 𝑥 ) = − csc ⁡ 2 𝑥 dx d ​ (cotx)=−csc 2 x: Solution: Start with the definition of cot ⁡ 𝑥 cotx: cot ⁡ 𝑥 = cos ⁡ 𝑥 sin ⁡ 𝑥 . cotx= sinx cosx ​ . Differentiate using the quotient rule: 𝑑 𝑑 𝑥 ( cot ⁡ 𝑥 ) = sin ⁡ 𝑥 ⋅ 𝑑 𝑑 𝑥 ( cos ⁡ 𝑥 ) − cos ⁡ 𝑥 ⋅ 𝑑 𝑑 𝑥 ( sin ⁡ 𝑥 ) ( sin ⁡ 𝑥 ) 2 . dx d ​ (cotx)= (sinx) 2 sinx⋅ dx d ​ (cosx)−cosx⋅ dx d ​ (sinx) ​ . Substitute 𝑑 𝑑 𝑥 ( cos ⁡ 𝑥 ) = − sin ⁡ 𝑥 dx d ​ (cosx)=−sinx and 𝑑 𝑑 𝑥 ( sin ⁡ 𝑥 ) = cos ⁡ 𝑥 dx d ​ (sinx)=cosx: 𝑑 𝑑 𝑥 ( cot ⁡ 𝑥 ) = sin ⁡ 𝑥 ( − sin ⁡ 𝑥 ) − cos ⁡ 𝑥 ( cos ⁡ 𝑥 ) ( sin ⁡ 𝑥 ) 2 . dx d ​ (cotx)= (sinx) 2 sinx(−sinx)−cosx(cosx) ​ . Simplify the numerator: 𝑑 𝑑 𝑥 ( cot ⁡ 𝑥 ) = − sin ⁡ 2 𝑥 − cos ⁡ 2 𝑥 sin ⁡ 2 𝑥 . dx d ​ (cotx)= sin 2 x −sin 2 x−cos 2 x ​ . Use the Pythagorean identity sin ⁡ 2 𝑥 + cos ⁡ 2 𝑥 = 1 sin 2 x+cos 2 x=1: 𝑑 𝑑 𝑥 ( cot ⁡ 𝑥 ) = − 1 sin ⁡ 2 𝑥 . dx d ​ (cotx)= sin 2 x −1 ​ . Recognize that csc ⁡ 𝑥 = 1 sin ⁡ 𝑥 cscx= sinx 1 ​ , so csc ⁡ 2 𝑥 = 1 sin ⁡ 2 𝑥 csc 2 x= sin 2 x 1 ​ : 𝑑 𝑑 𝑥 ( cot ⁡ 𝑥 ) = − csc ⁡ 2 𝑥 . dx d ​ (cotx)=−csc 2 x. Question 4 Find the derivative of 𝑠 ( 𝑡 ) = cos ⁡ 2 𝑡 s(t)=cos 2 t: Solution: Rewrite cos ⁡ 2 𝑡 cos 2 t as ( cos ⁡ 𝑡 ) 2 (cost) 2 . Differentiate using the chain rule: 𝑠 ′ ( 𝑡 ) = 2 cos ⁡ 𝑡 ⋅ 𝑑 𝑑 𝑡 ( cos ⁡ 𝑡 ) . s ′ (t)=2cost⋅ dt d ​ (cost). The derivative of cos ⁡ 𝑡 cost is − sin ⁡ 𝑡 −sint: 𝑠 ′ ( 𝑡 ) = 2 cos ⁡ 𝑡 ⋅ ( − sin ⁡ 𝑡 ) . s ′ (t)=2cost⋅(−sint). Simplify: 𝑠 ′ ( 𝑡 ) = − 2 cos ⁡ 𝑡 sin ⁡ 𝑡 . s ′ (t)=−2costsint. Use the double-angle identity sin ⁡ 2 𝑡 = 2 sin ⁡ 𝑡 cos ⁡ 𝑡 sin2t=2sintcost: 𝑠 ′ ( 𝑡 ) = − sin ⁡ 2 𝑡 . s ′ (t)=−sin2t. Question 5 Find the first and second derivatives of 𝑓 ( 𝑥 ) = ln ⁡ ( 𝑥 + 4 ) f(x)=ln(x+4): Solution: First derivative: Use the derivative of ln ⁡ 𝑢 lnu: 𝑑 𝑑 𝑥 ( ln ⁡ 𝑢 ) = 1 𝑢 ⋅ 𝑑 𝑢 𝑑 𝑥 dx d ​ (lnu)= u 1 ​ ⋅ dx du ​ . 𝑓 ′ ( 𝑥 ) = 1 𝑥 + 4 ⋅ 𝑑 𝑑 𝑥 ( 𝑥 + 4 ) . f ′ (x)= x+4 1 ​ ⋅ dx d ​ (x+4). Since 𝑑 𝑑 𝑥 ( 𝑥 + 4 ) = 1 dx d ​ (x+4)=1: 𝑓 ′ ( 𝑥 ) = 1 𝑥 + 4 . f ′ (x)= x+4 1 ​ . Second derivative: Differentiate 𝑓 ′ ( 𝑥 ) = 1 𝑥 + 4 f ′ (x)= x+4 1 ​ : Use the chain rule for ( 𝑥 + 4 ) − 1 (x+4) −1 : 𝑓 ′ ′ ( 𝑥 ) = − 1 ( 𝑥 + 4 ) 2 . f ′′ (x)=− (x+4) 2 1 ​ . Final Answer: First derivative: 𝑓 ′ ( 𝑥 ) = 1 𝑥 + 4 f ′ (x)= x+4 1 ​ . Second derivative: 𝑓 ′ ′ ( 𝑥 ) = − 1 ( 𝑥 + 4 ) 2 . f ′′ (x)=− (x+4) 2 1 ​ .