Questions and Answers #38 Derivatives

Questions and Answers Sheet 38 Derivatives Question #1 1 Find the derivative of 𝑒 𝑥 Answer: Use the chain rule: 𝑑 𝑓(𝑥) (𝑒 ) = 𝑒 𝑓(𝑥) × 𝑓 ′ (𝑥) 𝑑𝑥 The power of e is 1 𝑥 1 𝑥 1 , you should multiply 𝑒 𝑥 by the derivative of = 𝑥 −1 , its derivative is −𝑥 −2 = − 1 𝑥 1 𝑥2 1 1 1 𝑑 1 −𝑒 2 (𝑒 𝑥 ) = 𝑒 𝑥 × (− 2 ) = 2 𝑑𝑥 𝑥 𝑥 Question #2 Find the antiderivative of 𝑒 𝑥 2 Answer: 2 ∫ 𝑒 𝑥 𝑑𝑥 2 Use the nonelementary integral: ∫ 𝑒 𝑥 𝑑𝑥 = Add a constant √π 𝑒𝑟𝑓𝑖(𝑥) 2 √π 𝑒𝑟𝑓𝑖(𝑥) 2 +𝐶 Question #3 Find the derivative of 𝑥 2 𝑆 Answer: 𝑑 1 1 1 1 1 [ 𝑥 ] = 1 ( ) 𝑥 1−1 = 𝑥 0 = 𝑑𝑥 2 2 2 2 Question #4 Each limit represents the derivative of some function f at some number a. State such an f and a in each case. √9 + ℎ − 3 ℎ→0 ℎ lim Answer: Remember that: 𝑓 ′ (𝑎) = lim ℎ→0 𝑓(𝑎 + ℎ) − 𝑓(𝑎) ℎ √9 + ℎ − 3 ℎ→0 ℎ lim Re-write 3 in the numerator as √9 lim √9 + ℎ − √9 ℎ lim 𝑓(9 + ℎ) − 𝑓(9) = 𝑓 ′ (9) ℎ ℎ→0 ℎ→0 Where 𝑓(𝑥) = √𝑥 Note that 𝑎 = 9 Question #5 Find all the second partial derivatives. 𝑣 = 𝑥𝑦 𝑥−𝑦 Answer: 𝑣= 𝑥𝑦 𝑥−𝑦 Partially differentiate with respect to x (use the quotient rule) 𝑣𝑥 = [𝑑(𝑥𝑦)/𝑑𝑥] ⋅ (𝑥 − 𝑦) − (𝑥𝑦) ⋅ [𝑑(𝑥 − 𝑦)/𝑑𝑥] (𝑥 − 𝑦)2 = 𝑦(𝑥 − 𝑦) − (𝑥𝑦) ⋅ (1 − 0) (𝑥 − 𝑦)2 = 𝑥𝑦 − 𝑦 2 − 𝑥𝑦 −𝑦 2 = 2 (𝑥 − 𝑦) (𝑥 − 𝑦)2 Partially differentiate with respect to y (use the quotient rule) 𝑣𝑦 = [𝑑(𝑥𝑦)/𝑑𝑦] ⋅ (𝑥 − 𝑦) − (𝑥𝑦) ⋅ [𝑑(𝑥 − 𝑦)/𝑑𝑦] (𝑥 − 𝑦)2 = (𝑥)(𝑥 − 𝑦) − (𝑥𝑦) ⋅ (0 − 1) (𝑥 − 𝑦)2 = 𝑥 2 − 𝑥𝑦 + 𝑥𝑦 𝑥2 = 2 (𝑥 − 𝑦) (𝑥 − 𝑦)2 Partially differentiate 𝑣𝑥 with respect to x 𝑣𝑥𝑥 = = [𝑑(−𝑦 2 )/𝑑𝑥] ⋅ (𝑥 − 𝑦)2 − (−𝑦 2 ) ⋅ [𝑑(𝑥 − 𝑦)2 /𝑑𝑥] (𝑥 − 𝑦)4 0(𝑥 − 𝑦) − (−𝑦 2 ) ⋅ 2(𝑥 − 𝑦) ⋅ (1 − 0) (𝑥 − 𝑦)4 Cancel (𝑥 − 𝑦) from both the numerator and the denominator = 2𝑦 2 (𝑥 − 𝑦)3 Note that: If 𝑤 = 𝑦𝑥 𝑦−𝑥 2𝑥 2 , then we can write 𝑤𝑦 𝑦 = (𝑦−𝑥)3 But 𝑤 = −𝑣 , therefore 𝑣𝑦𝑦 = −𝑤𝑦𝑦 = 2𝑥 2 (𝑥 − 𝑦)3 Partially differentiate 𝑣𝑥 in with respect to y 𝑣𝑥𝑦 = = [𝑑(−𝑦 2 )/𝑑𝑦] ⋅ (𝑥 − 𝑦)2 − (−𝑦 2 ) ⋅ [𝑑(𝑥 − 𝑦)2 /𝑑𝑦] (𝑥 − 𝑦)4 (−2𝑦)(𝑥 − 𝑦)2 − (−𝑦 2 ) ⋅ 2 ⋅ (𝑥 − 𝑦) ⋅ (0 − 1) (𝑥 − 𝑦)4 Cancel (𝑥 − 𝑦) from both the numerator and the denominator = −2𝑦(𝑥 − 𝑦) − 2𝑦 2 −2𝑥𝑦 = (𝑥 − 𝑦)3 (𝑥 − 𝑦)3 Using the Clairaut's Theorem we can say that 𝑣𝑥𝑦 = 𝑣𝑦𝑥 = −2𝑥𝑦 (𝑥 − 𝑦)3 Question #6 What is the antiderivative of 𝑒 𝑥 ? Answer: Since (𝑒 𝑥 )′ = 𝑒 𝑥 we have ∫ 𝑒 𝑥 𝑑𝑥 = 𝑒 𝑥 + 𝐶 Question #7 How do you find the antiderivative of cos 2(𝑥) ? Answer: The trick to finding this integral is using an identity - here, specifically, the cosine double-angle identity. Since cos(2𝑥) = cos 2 (𝑥) − sin2 (𝑥) . We can rewrite this using the Pythagorean Identity to say that cos(2𝑥) = 2 cos 2 (𝑥) − 1 . Solving this for cos 2 (𝑥) shows us that cos 2 (𝑥) = 1 Thus, ∫ cos 2 (𝑥) 𝑑𝑥 = ∫ cos(2𝑥) + 1𝑑𝑥 2 We can now split this up and find the antiderivative. 1 1 = ∫ cos(2𝑥) 𝑑𝑥 + ∈ 1𝑑𝑥 2 2 1 1 = ∫ 2 cos(2𝑥) 𝑑𝑥 + 𝑥 4 2 1 1 = sin(2𝑥) + 𝑥 + 𝐶 4 2 Question #8 What is the derivative of 5𝑥 ? cos(2𝑥)+1 2 Answer: Explanation: Given: 𝑑 𝑑𝑥 (5𝑥 ) Apply the general formula that ∴ 𝑑 𝑑𝑥 (𝑎 𝑥 ) = 𝑎 𝑥 ln 𝑎 for ∈ 𝑅 𝑑 𝑥 (5 ) = 5𝑥 ln 5 𝑑𝑥 Question #9 If 𝑥 2 + 𝑥𝑦 + 𝑦 3 = 1 find the value of y''' at the point where 𝑥 = 1 Answer: Derivative with respect to 'x' 2𝑥 + 1 ⋅ 𝑦 + 𝑥 ⋅ 𝑦 ′ + 3𝑦 2 𝑦 ′ = 0 ⇒ 𝑦 ′ (𝑥 + 3𝑦 2 ) = −(2𝑥 + 𝑦) ⇒ 𝑦′ = −(2𝑥 + 𝑦) 𝑥 + 3𝑦 2 Now: Derivating with 'x' we get 2 + 𝑦 ′ + 𝑦 ′ + 𝑥𝑦 ′′ + 6𝑦(𝑦 ′ )2 + 3𝑦 2 𝑦 ′′ = 0 2 + 2𝑦 ′ + 6𝑦(𝑦 ′ )2 + (𝑥 + 3𝑦 2 )𝑦 ′′ = 0 derivative it one more time wait 'x' 2𝑦 ′′ + 6(𝑦 ′ )3 + 12𝑦𝑦 ′ 𝑦 ′′ + 𝑦 ′′ + 𝑥𝑦 ′′′ + 6𝑦𝑦 ′ 𝑦 ′′ + 3𝑦 2 𝑦 ′′′ = 0 𝑦 ′ 𝑎𝑡𝑥 = 1 ⇒ 1 + 𝑦 + 𝑦 3 = 1 𝑦(1 + 𝑦 2 ) = 0 𝑦=0 𝑦 ′ 𝑎𝑡𝑥 = 1 ⇒ 𝑦 ′ = −(2 ⋅ +0) 1+2⋅0 𝑦 ′ = −2 𝑦 ′′ 𝑎𝑡𝑥 = 1 ⇒ 2 + 2(−2) + 6𝑡(0)(−2)2 + (1 + 3(0)2 )𝑦 ′′ = 0 𝑦 ′′ = 2 𝑦 ′′′ 𝑎𝑡𝑥 = 1 ⇒ 2(2) + 6(−2)3 + 18(0)(−2)(2) + (+2) + (1 + 3(0))𝑦 ′′′ = 0 𝑦 ′′′ 𝑎𝑡𝑥 = 1 ⇒ 2(2) + 6(−2)3 + 18(0)(−2)(2) + (+2) + (1 + 3(0))𝑦 ′′′ = 0 4 − 48 + 2 + 𝑦 ′′′ = 0 𝑦 ′′′ = 42 Question #10 What is the derivative of 𝑦 = arctan(𝑥) ? Answer: The derivative of 𝑦 = arctan 𝑥 is 𝑦 ′ = 1 1+𝑥 2 We can derive this by using implicit differentiation. Since inverse tangent is hard to deal with, we rewrite it as tan(𝑦) = 𝑥 By implicitly differentiating with respect to x, sec 2 (𝑦) ⋅ 𝑦 ′ = 1 By solbing for y' and using 𝑠𝑒𝑐 2 (𝑦) = 1 + tan2 (𝑦) 𝑦′ = 1 1 = sec 2 (𝑦) tan2 (𝑦) Hence, 𝑦 ′ = 1 1+𝑥 2 Question #11 What is the derivative of 3𝑥 ? Answer: Begin by letting 𝑦 = 3𝑥 now take the ln of both sides. ln 𝑦 = ln 3𝑥 ⇒ ln 𝑦 = 𝑥 ln 3 differentiate implicitly with respect to x ⇒ 1 𝑑𝑦 = ln 3 𝑦 𝑑𝑥 Now 𝑦 = 3𝑥 ⇒ 𝑑𝑦 𝑑𝑥 = 3𝑥 ln 3 This result can be generalised as follows 𝑑 𝑥 (𝑎 ) = 𝑎 𝑥 ln 𝑎 𝑑𝑥 Question #12 How do you find the derivative of cot 𝑥 ? Answer: Explanation: 𝑦 = cot 𝑥 𝑦= 1 tan 𝑥 1 sin 𝑥 cos 𝑥 cos 𝑥 𝑦= sin 𝑥 𝑦= Letting 𝑦 = 𝑦′ = 𝑔(𝑥) ℎ(𝑥) , we have that 𝑔(𝑥) = cos 𝑥 and ℎ(𝑥) = sin 𝑥 𝑔′ (𝑥) × ℎ(𝑥) − 𝑔(𝑥) × ℎ′ (𝑥) (ℎ(𝑥)) 2 𝑦′ = − sin 𝑥 × sin 𝑥 − (cos 𝑥 × cos 𝑥) (sin 𝑥)2 𝑦′ = − sin2 𝑥 + cos 2 𝑥 sin2 𝑥 𝑦′ = −(sin2 𝑥 + cos 2 𝑥) sin2 𝑥 𝑦′ = − 1 sin2 𝑥 𝑦 ′ = − csc 2 𝑥 Question #13 What is the derivative of tan(2𝑥) ? Answer: Assuming that you know the derivative rule: 𝑑 𝑑𝑥 (tan(2𝑥)) will simply be sec 2 (2𝑥) ⋅ Then 𝑑 𝑑𝑥 𝑑 𝑑𝑥 𝑑 𝑑𝑥 (tan 𝑥) = sec 2 (𝑥) (2𝑥) according to the chain rule. (tan(2𝑥)) = 2 sec 2 (2𝑥) If you want to easily understand chain rule, just remember my tips: take the normal derivative of the outside (ignoring whatever is inside the parenthesis) and then multiply it by the derivative of the inside (stuff inside the parenthesis) Question #14 Solve the following initial value problem for the function y(x). 𝑦 ′ = 2𝑥𝑦 2 ; 𝑦(1) = Answer: 𝑦 ′ = 2𝑥𝑦 2 ⇒ 𝑑𝑦 = 2𝑥𝑦 2 𝑑𝑥 ⇒∫ 𝑑𝑦 = ∫ 2𝑥𝑑𝑥 𝑦2 = −𝑦 −1 = 𝑥 2 + 𝑐 ⇒ −1 = 𝑥2 + 𝑐 𝑦 given 𝑦(1) = 1 2 ⇒ −2 = 1 + 𝑐 1 2 ⇒ 𝑐 = −3 ∴ −1 = 𝑥2 − 3 𝑦 1 ⇒ 𝑦 = − 𝑥 2 - is the solution 3 Question #15 Find all the second partial derivatives 𝑓(𝑥, 𝑦) = 𝑠𝑖𝑛(2) (𝑚𝑥 + 𝑛𝑦) Answer: 𝑓(𝑥, 𝑦) = 𝑠𝑖𝑛(2) (𝑚𝑥 + 𝑛𝑦) 𝑓𝑥 = 2 sin(𝑚𝑥 + 𝑛𝑦) × 𝜕 [sin(𝑚𝑥 + 𝑛𝑦)] 𝜕𝑥 = 2𝑚 sin(𝑚𝑥 + 𝑛𝑦) × cos(𝑚𝑥 + 𝑛𝑦) Similary 𝑓𝑦 = 2𝑛 sin(𝑚𝑥 + 𝑛𝑦) × cos(𝑚𝑥 + 𝑛𝑦) ⇒ 𝑓𝑥 = 𝑚 sin(2(𝑚𝑥 + 𝑛𝑦)) 𝑓𝑦 = 𝑛 sin(2(𝑚𝑥 + 𝑛𝑦)) 𝑓(𝑥𝑥) = 𝑚 ∗ 𝜕 sin(2(𝑚𝑥 + 𝑛𝑦)) 𝜕𝑥 = 2𝑚 × 𝑚 × cos(2(𝑚𝑥 + 𝑛𝑦)) 𝑓(𝑥𝑥) = 2𝑚2 cos(2(𝑚𝑥 + 𝑛𝑦)) 𝑓(𝑦𝑦) = 𝑛 ∗ 𝜕 sin(2(𝑚𝑥 + 𝑛𝑦)) 𝜕𝑦 = 2𝑛 × 𝑛 × cos(2(𝑚𝑥 + 𝑛𝑦)) 𝑓(𝑦𝑦) = 2𝑛2 cos(2(𝑚𝑥 + 𝑛𝑦)) Question #16 Find the derivative of − sin(𝑥) Answer: sin(𝑥 + ℎ) − sin(𝑥) ℎ→0 ℎ 𝑓 ′ (𝑥) = lim Let's use representation of a difference of sin functions as a product of sin and cos, and we get: 𝑓 ′ (𝑥) ℎ ℎ 2 × sin ( ) cos (𝑥 + ) 2 2 = lim ℎ→0 ℎ ℎ sin ( ) 2 × lim cos (𝑥 + ℎ) ℎ ℎ→0 ℎ→0 2 2 𝑓 ′ (𝑥) = lim 𝑓 ′ (𝑥) = 1 × cos(𝑥) = cos(𝑥) Thus, the derivative is 𝑓 ′ (𝑥) = cos(𝑥) Question #17 Find the antiderivative of csc 2 𝑥 Answer: 𝑑 (cot(𝑥)) = − csc 2 (𝑥) 𝑑𝑥 𝑑 (− cot(𝑥)) = −(− csc 2 (𝑥)) 𝑑𝑥 = csc 2 (𝑥) − cot(𝑥) + 𝐶 is the antiderivative Question #18 How many tangent lines to the curve 𝑦 = 𝑥 𝑥+1 pass through the point (1,2) Answer: The form of the tangent lines must be: 𝑦 = 𝑚(𝑥 − 1) + 2[1] Now, substitute the first derivative of the curve for m 𝑑(𝑥) 𝑑(𝑥 + 1) (𝑥 + 1) − 𝑥 𝑑𝑦 𝑑𝑥 𝑑𝑥 = (𝑥 + 1)2 𝑑𝑥 𝑑𝑦 𝑥 + 1 − 𝑥 = (𝑥 + 1)2 𝑑𝑥 𝑑𝑦 1 = 𝑑𝑥 (𝑥 + 1)2 𝑚= 1 [2] (𝑥 + 1)2 Now, substitute equation [2] into equation [1] 𝑦= 𝑥−1 + 2[1.1] (𝑥 + 1)2 We have to substitute 𝑦 = 𝑥 𝑥+1 𝑥 𝑥−1 = +2 𝑥 + 1 (𝑥 + 1)2 Solve for x 𝑥(𝑥 + 1) = (𝑥 − 1) + 2(𝑥 + 1)2 𝑥 2 + 𝑥 = 𝑥 − 1 + 2𝑥 2 + 4𝑥 + 2 𝑥 2 + 4𝑥 + 1 𝑥= −4 ± √42 − 4(1)(1) 2(1) 𝑥 = −2 ± √3 𝑥 = −2 + √3 and 𝑥 = −2 − √3 There are two tangent lines 𝑦= 1 2 (−1+√3) (𝑥 − 1) + 2 and 𝑦 = 1 2 (−1−√3) (𝑥 − 1) + 2 Question #19 What is the derivative of sin−1 (𝑥) Answer: Let 𝑦 = sin−1 (𝑥) 𝜋 𝜋 2 2 Thus, sin(𝑦) = 𝑥 and − ≤ 𝑦 ≤ Now, diferentiate implicity cos(𝑦) 𝜋 𝜋 2 2 𝑑𝑦 𝑑𝑥 = 1 , so 𝑑𝑦 𝑑𝑥 = 1 cos(𝑦) As we know, that − ≤ 𝑦 ≤ , we understand that cos(𝑦) is positive 𝑑𝑦 1 1 = = 𝑑𝑥 √1 − sin2 (𝑦) √1 − 𝑥 2 Question #20 Compute the first-order partial derivatives. 𝑧 = 𝑥 2 + 𝑦2 Answer: Step 1 Explanation: Given that, 𝑧 = 𝑥 2 + 𝑦2 Concept used: Here we treat y as a constant. Sum and difference rule of derivatives: (𝑓 ± 𝑔)′ = 𝑓 ′ ± 𝑔′ Step 2 Calculation: 𝑧 = 𝑥 2 + 𝑦2 𝜕𝑧 𝜕 2 (𝑥 + 𝑦 2 ) = 𝜕𝑥 𝜕𝑥 = 𝜕 2 𝜕 2 (𝑥 ) + (𝑦 ) 𝜕𝑥 𝜕𝑥 = 2𝑥 + 0 = 2𝑥 Hence, the first-order partial derivative of 𝑧 = 𝑥 2 + 𝑦 2 is 2x