Derivatives #13 - Questions and Answers

Question 1 Calculate each function’s gradient and evaluate it at 𝑃 = ( − 1 , 2 ) P=(−1,2). Answer: Given the function: 𝑔 ( 𝑥 , 𝑦 ) = 𝑥 2 − 4 𝑥 2 𝑦 − 8 𝑥 𝑦 2 g(x,y)=x 2 −4x 2 y−8xy 2 The gradient is the vector of partial derivatives: ∇ 𝑔 = ∂ 𝑔 ∂ 𝑥 𝑖 + ∂ 𝑔 ∂ 𝑦 𝑗 ∇g= ∂x ∂g ​ i+ ∂y ∂g ​ j Partial derivative with respect to 𝑥 x: ∂ 𝑔 ∂ 𝑥 = 2 𝑥 − 8 𝑥 𝑦 − 8 𝑦 2 ∂x ∂g ​ =2x−8xy−8y 2 At 𝑃 ( − 1 , 2 ) P(−1,2): ∂ 𝑔 ∂ 𝑥 = 2 ( − 1 ) − 8 ( − 1 ) ( 2 ) − 8 ( 2 ) 2 = − 2 + 16 − 32 = − 18. ∂x ∂g ​ =2(−1)−8(−1)(2)−8(2) 2 =−2+16−32=−18. Partial derivative with respect to 𝑦 y: ∂ 𝑔 ∂ 𝑦 = − 4 𝑥 2 − 16 𝑥 𝑦 ∂y ∂g ​ =−4x 2 −16xy At 𝑃 ( − 1 , 2 ) P(−1,2): ∂ 𝑔 ∂ 𝑦 = − 4 ( − 1 ) 2 − 16 ( − 1 ) ( 2 ) = − 4 + 32 = 28. ∂y ∂g ​ =−4(−1) 2 −16(−1)(2)=−4+32=28. Gradient at 𝑃 ( − 1 , 2 ) P(−1,2): ∇ 𝑔 = − 18 𝑖 + 28 𝑗 . ∇g=−18i+28j. Question 2 Find the derivative of 𝑦 = 2 tan ⁡ ( 5 𝑥 ) y=2tan(5x). Answer: Differentiate using the chain rule: 𝑑 𝑑 𝑥 ( tan ⁡ ( 5 𝑥 ) ) = sec ⁡ 2 ( 5 𝑥 ) ⋅ 𝑑 𝑑 𝑥 ( 5 𝑥 ) = 5 sec ⁡ 2 ( 5 𝑥 ) . dx d ​ (tan(5x))=sec 2 (5x)⋅ dx d ​ (5x)=5sec 2 (5x). Multiply by the constant 2 2: 𝑑 𝑦 𝑑 𝑥 = 2 ⋅ 5 sec ⁡ 2 ( 5 𝑥 ) = 10 sec ⁡ 2 ( 5 𝑥 ) . dx dy ​ =2⋅5sec 2 (5x)=10sec 2 (5x). Question 3 Use the product rule twice to find a formula for ( 𝑓 𝑔 ) ′ (fg) ′ in terms of 𝑓 f and 𝑔 g and their derivatives. Answer: Start with the product rule: ( 𝑓 𝑔 ) ′ = 𝑓 ′ 𝑔 + 𝑓 𝑔 ′ . (fg) ′ =f ′ g+fg ′ . Differentiate again to find the second derivative: ( 𝑓 𝑔 ) ′ ′ = 𝑑 𝑑 𝑥 ( 𝑓 ′ 𝑔 + 𝑓 𝑔 ′ ) = 𝑓 ′ ′ 𝑔 + 𝑓 ′ 𝑔 ′ + 𝑓 ′ 𝑔 ′ + 𝑓 𝑔 ′ ′ . (fg) ′′ = dx d ​ (f ′ g+fg ′ )=f ′′ g+f ′ g ′ +f ′ g ′ +fg ′′ . Combine terms: ( 𝑓 𝑔 ) ′ ′ = 𝑓 ′ ′ 𝑔 + 2 𝑓 ′ 𝑔 ′ + 𝑓 𝑔 ′ ′ . (fg) ′′ =f ′′ g+2f ′ g ′ +fg ′′ . Thus, the formula for the first and second derivatives of 𝑓 𝑔 fg is: ( 𝑓 𝑔 ) ′ = 𝑓 ′ 𝑔 + 𝑓 𝑔 ′ , ( 𝑓 𝑔 ) ′ ′ = 𝑓 ′ ′ 𝑔 + 2 𝑓 ′ 𝑔 ′ + 𝑓 𝑔 ′ ′ . (fg) ′ =f ′ g+fg ′ ,(fg) ′′ =f ′′ g+2f ′ g ′ +fg ′′ . Question 4 Find the derivative of 𝑔 ( 𝑤 ) = 𝑒 𝑤 + 𝑒 − 𝑤 𝑒 𝑤 g(w)= e w e w +e −w ​ . Answer: Simplify the given function: 𝑔 ( 𝑤 ) = 𝑒 𝑤 + 𝑒 − 𝑤 𝑒 𝑤 = 1 + 𝑒 − 2 𝑤 . g(w)= e w e w +e −w ​ =1+e −2w . Differentiate with respect to 𝑤 w: 𝑔 ′ ( 𝑤 ) = 𝑑 𝑑 𝑤 [ 1 ] + 𝑑 𝑑 𝑤 [ 𝑒 − 2 𝑤 ] . g ′ (w)= dw d ​ [1]+ dw d ​ [e −2w ]. The derivative of 1 1 is 0 0. Differentiate 𝑒 − 2 𝑤 e −2w : 𝑑 𝑑 𝑤 [ 𝑒 − 2 𝑤 ] = − 2 𝑒 − 2 𝑤 . dw d ​ [e −2w ]=−2e −2w . Combine results: 𝑔 ′ ( 𝑤 ) = − 2 𝑒 − 2 𝑤 . g ′ (w)=−2e −2w . Question 5 Estimate the slopes 𝑓 ′ ( 𝑥 ) f ′ (x) at specific points using the graph. Answer: (a) 𝑓 ′ ( 0 ) f ′ (0): From the graph, the slope at 𝑥 = 0 x=0 is: 𝑓 ′ ( 0 ) = rise run = − 3 1 = − 3. f ′ (0)= run rise ​ = 1 −3 ​ =−3. (b) 𝑓 ′ ( 1 ) f ′ (1): The tangent line is horizontal at 𝑥 = 1 x=1, so: 𝑓 ′ ( 1 ) = 0. f ′ (1)=0. (c) 𝑓 ′ ( 2 ) f ′ (2): From the graph, the slope at 𝑥 = 2 x=2: 𝑓 ′ ( 2 ) = rise run = 2 5 = 0.4. f ′ (2)= run rise ​ = 5 2 ​ =0.4. (d) 𝑓 ′ ( 3 ) f ′ (3): From the graph, the slope at 𝑥 = 3 x=3: 𝑓 ′ ( 3 ) = rise run = 2 1 = 2. f ′ (3)= run rise ​ = 1 2 ​ =2. (e) 𝑓 ′ ( 4 ) f ′ (4): The tangent line is horizontal at 𝑥 = 4 x=4, so: 𝑓 ′ ( 4 ) = 0. f ′ (4)=0. (f) 𝑓 ′ ( 5 ) f ′ (5): From the graph, the slope at 𝑥 = 5 x=5: 𝑓 ′ ( 5 ) = rise run = − 1.5 1.5 = − 1. f ′ (5)= run rise ​ = 1.5 −1.5 ​ =−1. Summary of Slopes: 𝑓 ′ ( 0 ) = − 3 f ′ (0)=−3 𝑓 ′ ( 1 ) = 0 f ′ (1)=0 𝑓 ′ ( 2 ) = 0.4 f ′ (2)=0.4 𝑓 ′ ( 3 ) = 2 f ′ (3)=2 𝑓 ′ ( 4 ) = 0 f ′ (4)=0 𝑓 ′ ( 5 ) = − 1 f ′ (5)=−1.