Questions and Answers #7 Derivatives

Questions and answers Sheet #7 Derivatives Question #1 Find the following derivatives. 𝑧𝑠 and 𝑧𝑡 , where 𝑧 = 𝑒 𝑥+𝑦 , 𝑥 = 𝑠𝑡, and 𝑦 = 𝑠 + 𝑡𝑎𝑠𝑘 Answer: Given The given eqaution is 𝑧 = 𝑒 𝑥+𝑦 , 𝑥 = 𝑠𝑡 and 𝑦 = 𝑠 + 𝑡 Differential formula, 𝑑 𝑑𝑥 (𝑒 𝑎𝑥 ) = 𝑒 𝑎𝑥 . 𝑎 Solution The given eqaution is 𝑧 = 𝑒 𝑥+𝑦 , 𝑥 = 𝑠𝑡 and 𝑦 = 𝑠 + 𝑡 𝑧 = 𝑒 𝑠𝑡+𝑠+𝑡 Now, 𝑧𝑠 = 𝑧𝑡 = 𝑑𝑧 𝑑𝑠 = 𝑒 𝑠𝑡+𝑠+𝑡 (𝑡 + 1) 𝑑𝑧 = 𝑒 𝑠𝑡+𝑠+𝑡 (𝑠 + 1) 𝑑𝑡 Question #2 Find the first partial derivatives of the following functions. 𝑓(𝑥, 𝑦) = 4𝑥 3 𝑦 2 + 3𝑥 2 𝑦 3 + 10 Answer: Step 1 Given, 𝑓(𝑥, 𝑦) = 4𝑥 3 𝑦 2 + 3𝑥 2 𝑦 3 + 10 The partial derivative of f(x,y) with respect to x is, 𝑓 ′ (𝑥, 𝑦) = = 𝜕 (4𝑥 3 𝑦 2 + 3𝑥 2 𝑦 3 + 10) 𝜕𝑥 𝜕 𝜕 (4𝑥 3 𝑦 2 ) + (3𝑥 2 𝑦 3 + 10) 𝜕𝑥 𝜕𝑥 = 12𝑦 2 𝑥 2 + 𝑦 3 𝑥 + 0 = 12𝑦 2 𝑥 2 + 𝑦 3 𝑥 Step 2 Similarly, The partial derivative of f(x,y) with respect to y is 𝑓 ′ (𝑥, 𝑦) = = 𝜕 𝜕 (3𝑥 2 𝑦 3 ) + (10) 𝜕𝑦 𝜕𝑦 = 8𝑥 3 𝑦 + 9𝑥 2 𝑦 2 + 0 𝜕 𝜕𝑦 (4𝑥 3 𝑦 2 + 3𝑥 2 𝑦 3 + 10) = 8𝑥 3 𝑦 + 9𝑥 2 𝑦 2 Question #3 Find the derivatives of the following function with respect to x and simplify the result. 𝑦 = cos ℎ (log 𝑥) Answer: Step 1 To find the derivative of hyperbolic function with respect to x Use chain rule Step 2 𝑦 = cos ℎ (log 𝑥) 𝑑𝑦 𝑑 = (cos ℎ (log 𝑥)) 𝑑𝑥 𝑑𝑥 = sin ℎ (log 𝑥) ∗ 𝑑 (log 𝑥) 𝑑𝑥 1 = sin ℎ (log 𝑥) ∗ ( ) 𝑥 Question #4 Find the derivatives of the following function with respect to x and simplify the result. 𝑦 = 𝑒 cot ℎ −1 𝑥 Answer: Step 1 Ans: 𝑦 = 𝑒 cot ℎ ⇒ = −1 𝑥 𝑑 cot ℎ−1(𝑥) 𝑑 1 −1 −1 [cot ℎ−1 (𝑥)] = 𝑒 cot ℎ (𝑥) ∗ [𝑒 ] = 𝑒 cot ℎ (𝑥) ∗ 𝑑𝑥 𝑑𝑥 1 − 𝑥2 𝑒 cot ℎ −1 (𝑥) 1−𝑥 2 or − 𝑒 cot ℎ −1 (𝑥) 𝑥 2 −1 Step 2 Is the required answer. Question #5 Given 𝑓(𝑥, 𝑦) = 2𝑥 2 − 𝑥𝑦 3 + 4𝑦 6 , find 𝑓𝑥𝑥 (𝑥, 𝑦) = 𝑓𝑥𝑦 (𝑥, 𝑦) = Answer: Step 1 Consider the given function. 𝑓(𝑥, 𝑦) = 2𝑥 2 − 𝑥𝑦 3 + 4𝑦 6 The objective of the question is to find the value of 𝑓𝑥𝑥 (𝑥, 𝑦) and 𝑓𝑥𝑦 (𝑥, 𝑦) . These are the second order partial derivatives. Step 2 Partially differentiate the function with respect to x. 𝑓𝑥 (𝑥, 𝑦) = 2 ∗ 3𝑥 3−1 − 1 ∗ 𝑦 3 + 0 = 6𝑥 2 − 𝑦 3 Partially differentiate again with respect to x to find 𝑓𝑥𝑥 (𝑥, 𝑦). 𝑓𝑥𝑥 (𝑥, 𝑦) = 6 ∗ 2𝑥 − 0 = 12𝑥 Partially differentiate the function with respect to y. 𝑓𝑦 (𝑥, 𝑦) = 0 − 𝑥 ∗ 3𝑦 3−1 + 4 ∗ 6𝑦 6−1 = −3𝑥𝑦 2 + 24𝑦 5 Partially differentiate again with respect to x to find 𝑓𝑥𝑦 (𝑥, 𝑦). 𝑓𝑥𝑦 (𝑥, 𝑦) = −3𝑦 2 ∗ 1 + 0 = −3𝑦 2 Thus, the required partial derivatives are obtained. 𝑓𝑥𝑥 (𝑥, 𝑦) = 12𝑥 𝑓𝑥𝑦 (𝑥, 𝑦) = −3𝑦 2 Question #6 Find the derivatives of the following functions with respect to x and simplify the result. 𝑓(𝑥) = ln cos ℎ (4𝑥) Answer: Step 1 The given function is, 𝑓(𝑥) = ln cos ℎ (4𝑥) Then, differentiate by the chain rule is, 𝑓(𝑥) = 𝑑 𝑑 𝑑 (ln cos ℎ (4𝑥)) [cos ℎ (4𝑥)] 4𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥 = 1 [4 sin ℎ (4𝑥)](4) cos ℎ (4𝑥) = 16 sin ℎ (4𝑥) cos ℎ (4𝑥) Step 2 Thus, the value of the differentiation is Question #7 Find the higher-order derivative. 𝑓 ′′ (𝑥) = 10𝑥 3 − 32𝑥 2 16 sin ℎ(4𝑥) cos ℎ(4𝑥) 𝑓 ′′′ (𝑥) = Answer: Step 1 This differentiation process can be continued to find the third, fourth, and successive derivatives of f( x), which are called higher order derivatives of f( x). Step 2 𝑓 ′′ (𝑥) = 10𝑥 3 − 32𝑥 2 𝑓 ′′′ (𝑥) = 30𝑥 2 − 64𝑥 𝑓 ′′′′ (𝑥) = 60𝑥 − 64 Question #8 Using the definition, calculate the derivatives of the functions. Then find the values of the derivatives as specified. 𝑘(𝑧) = 1−𝑧 ′ ; 𝑘 (−1), 𝑘 ′ (1), 𝑘 ′ (√2) 2𝑧 Answer: Step 1 𝑘(𝑧) = 1−𝑧 2𝑧 Step 2 Using Quotient Rule 𝑓 𝑓 ′ 𝑔 − 𝑔′ 𝑓 ( )= 𝑔 𝑔2 Step 3 Using quotient rule to find the derivative 𝑘 ′ (𝑧) = 1 1−𝑧 [ ] 2 𝑧 𝑘 ′ (𝑧) = 1 𝑧(1 − 𝑧)′ − 𝑧 ′ (1 − 𝑧) [ ] 2 𝑧2 𝑘 ′ (𝑧) = 1 −𝑧 − (1 − 𝑧) [ ] 2 𝑧2 𝑘 ′ (𝑧) = 1 −𝑧 − 1 + 𝑧 [ ] 2 𝑧2 1 𝑘 ′ (𝑧) = [ 2 ] 2𝑧 Step 4 Put 𝑧 = 1 1 𝑘 ′ (1) = [ ] 2 Put 𝑧 = −1 1 𝑘 ′ (−1) = [ ] 2 put 𝑧 = √2 1 𝑘 ′ (√2) = [ ] 4 Question #9 Find the derivatives of the functions. 𝑤 = 𝑟𝑒 −𝑟 Answer: Step 1 Given 𝑤 = 𝑟𝑒 −𝑟 Step 2 Find the derivatives of the function; 𝑑𝑤 𝑑 (𝑟𝑒 −𝑟 ) = 𝑑𝑟 𝑑𝑟 = 𝑒 −𝑟 𝑑 𝑑 (𝑟) + 𝑟 (𝑒 −𝑟 ) 𝑑𝑟 𝑑𝑟 (Apply product rule; (𝑓 ∗ 𝑔)′ = 𝑔𝑓′ + 𝑓𝑔′) = 𝑒 −𝑟 (1) + 𝑟(−𝑒 −𝑟 ) = 𝑒 −𝑟 (1 − 𝑟) Question #10 Evaluate the derivatives of the following functions. 𝑓(𝑠) = cot −1 (𝑒 𝑠 ) Answer: Step 1 The given function is 𝑓(𝑠) = cot −1 (𝑒 𝑠 ) Step 2 Evaluate the derivatives of 𝑓(𝑠) = cot −1 (𝑒 𝑠 ) 𝑓 ′ (𝑠) = =− =− =− 𝑑 (cot −1 (𝑒 𝑠 )) 𝑑𝑠 1 (𝑒 𝑠 )2 𝑑 𝑠 (𝑒 ) + 1 𝑑𝑠 1 (𝑒 𝑠 )2 +1 𝑒𝑠 1 + 𝑒 2𝑠 (𝑒 𝑠 ) Question #11 Find all derivatives of the function 𝑦 = 1 2𝑥−3 Answer: Step 1 We find first few derivatives using the power rule and chain rule 𝑦= 1 = (2𝑥 − 3)−1 2𝑥 − 3 𝑦 (1) = −1(2𝑥 − 3)−2 (2) = −2(2𝑥 − 3)−2 = (−1)(2)(1!)(2𝑥 − 3)−2 𝑦 (2) = 4(2𝑥 − 3)−3 (2) = 8(2𝑥 − 3)−3 = (−1)2 (2)(2!)(2𝑥 − 3)−3 𝑦 (3) = −12(2𝑥 − 3)−3 (2) = −24(2𝑥 − 3)−4 = (−1)3 (2)(3!)(2𝑥 − 3)−4 𝑦 (4) = 36(2𝑥 − 3)−3 (2) = 72(2𝑥 − 3)−5 = (−1)4 (2)(4!)(2𝑥 − 3)−5 Step 2 Then we find a pattern for integer n 𝑦 (𝑛) = (−1)𝑛 (2)(𝑛!)(2𝑥 − 3)−𝑛−1 Answer: 𝑦 (𝑛) = (−1)𝑛 (2)(𝑛!)(2𝑥 − 3)−𝑛−1 Question #12 Consider the following. 𝑓(𝑥) = 6𝑥 4 − 6𝑥 3 Find the following derivatives. 𝑓 ′ (𝑥) = 𝑓 ′′ (𝑥) = 𝑓 ′′′ (𝑥) = Answer: Step 1 From the given data: Step 2 𝑓(𝑥) = 6𝑥 4 − 6𝑥 3 Now, differentiating the above equation with respect to x 𝑓 ′ (𝑥) = 24𝑥 3 − 18𝑥 2 Now, differentiating the above equation with respect to x 𝑓 ′′ (𝑥) = 72𝑥 2 − 36𝑥 Now, differentiating the above equation with respect to x 𝑓 ′′′ (𝑥) = 144𝑥 − 36 Question #13 Find the derivatives of the functions. 𝑦 = 4 − 2𝑥 − 𝑥 −3 Answer: Step 1 Find the derivatives of the functions 𝑦 = 4 − 2𝑥 − 𝑥 −3 Step 2 𝑦 = 4 − 2𝑥 − 𝑥 −3 differentiate y with respect to x, we get 𝑑𝑦 = 0 − 2 − (−3)𝑥 −3−1 𝑑𝑥 ⇒ 𝑑𝑦 = −2 + 3𝑥 −2 𝑑𝑥 This is the required answer. Question #14 Let f and g be differentiable functions. Find a formula for 𝑑 𝑑𝑥 3𝑓(𝑥) [ 2 − 5𝑔(𝑥)] Answer: Step 1 Derivative of f(x) is f'(x). Derivative of g(x) is g'(x). Step 2 𝑑 3𝑓(𝑥) [ − 5𝑔(𝑥)] 𝑑𝑥 2 = 𝑑 3𝑓(𝑥) 𝑑 [5𝑔(𝑥)] [ ]− 𝑑𝑥 2 𝑑𝑥 = 3 𝑑 𝑑 [𝑓(𝑥)] − 5 [𝑔(𝑥)] 2 𝑑𝑥 𝑑𝑥 3 = 𝑓 ′ (𝑥) − 5𝑔′ (𝑥) 2 3 Answer: 𝑓 ′ (𝑥) − 5𝑔′ (𝑥) 2 Question #15 Using the definition, calculate the derivatives of the function. Then find the values of the derivatives as specified. 𝑔(𝑡) = 1 ′ ; 𝑔 (−1), 𝑔′ (2), 𝑔′ (√3) 𝑡2 Answer: Step 1 The given function is, 𝑔(𝑡) = 1 𝑡2 To find the derivative of the function, we use the power rule of differentiation, 𝑑 𝑛 (𝑥 ) = 𝑛𝑥 𝑛−1 𝑑𝑥 Step 2 Applying differentiation on both sides of the function, we get 𝑔′ (𝑡) = = 𝑑 1 ( ) 𝑑𝑡 𝑡 2 𝑑 −2 (𝑡 ) 𝑑𝑡 = −2𝑡 −2−1 = −2𝑡 −3 =− 2 𝑡3 Therefore, the derivative of the given function is 𝑔′ (𝑡) = − 2 𝑡3 Step 3 Finding the value of the derivative of the given function for the specified value of the variable, we get 𝑔′ (−1) = − =− 2 (−1)3 2 −1 =2 𝑔′ (2) = − =− 2 8 =− 1 4 2 (2)3 2 𝑔′ (√3) = − 3 (√3) =− 2 3√3 Question #16 3 𝜋3 5 2 Find the derivatives of the functions 𝑦 = 𝑥 − + Answer: Step 1 Given function is a linear function 3 𝜋3 𝑦=𝑥− + 5 2 𝑦′ = 1 𝑦 ′′ = 0 Step 2 Other higher-order derivatives will be 0 too Answer: 𝑦 = 1, 𝑦 (𝑛) = 0 for 𝑛 > 1 Question #17 Use the theorems on derivatives to find the derivatives of the following function: 𝑓(𝑥) = 3𝑥 5 − 2𝑥 4 − 5𝑥 + 7 + 4𝑥 −2 Answer: Step 1 There are two operations addition and substraction in the given function which is a polynomial. We use sum/difference rule of differentiation, (𝑓 ± 𝑔)′ = 𝑓 ′ ± 𝑔′ Also we know that the derivative of 𝑥 𝑎 is 𝑎𝑥 𝑎−1 . Derivative of a constant is always zero. Step 2 Given 𝑓(𝑥) = 3𝑥 5 − 2𝑥 4 − 5𝑥 + 7 + 4𝑥 −2 𝑓 ′ (𝑥) = 𝑑 𝑑 𝑑 𝑑 𝑑 (3𝑥 5 ) − (2𝑥 4 ) − (5𝑥) + (7) + (4𝑥 −2 ) 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥 = 15𝑥 4 − 8𝑥 3 − 5 + 0 − 8𝑥 3 Step 3 Therefore the derivative is given as, 𝑓 ′ (𝑥) = 15𝑥 4 − 8𝑥 3 − Question #18 Find the indicated derivatives. 𝑑𝑦 𝑑𝑥 if 𝑦 = 2𝑥 3 Answer: 𝑦 = 2𝑥 3 Differentiating with respect to x we get 𝑑𝑦 𝑑 (2𝑥 3 ) = 𝑑𝑥 𝑑𝑥 𝑑𝑦 𝑑 = 2 (𝑥 3 ) 𝑑𝑥 𝑑𝑥 We know that 𝑑 𝑑𝑥 (𝑥 𝑛 ) = 𝑛𝑥 𝑛−1 8 𝑥3 −5 ⇒ 𝑑𝑦 = 2(3𝑥 2 ) 𝑑𝑥 Hence 𝑑𝑦 𝑑𝑥 = 6𝑥 2 Question #19 Find the indicated derivatives. 𝑟 = 𝑠 3 − 2𝑠 2 + 3 Answer: 𝑟 = 𝑠 3 − 2𝑠 2 + 3 Differentiating with respect to s we get 𝑑𝑟 𝑑 3 (𝑠 − 2𝑠 2 + 3) = 𝑑𝑠 𝑑𝑠 𝑑𝑟 = 3(𝑠 3−1 ) − 2(2𝑠 2−1 ) + 0 𝑑𝑠 𝑑𝑟 = 3𝑠 2 − 2(2𝑠) 𝑑𝑠 Hence 𝑑𝑟 𝑑𝑠 = 3𝑠 2 − 4𝑠 Question #20 How are the derivatives of sin−1 𝑥 and cos −1 𝑥 related? Answer: Step 1 Given functions are sin−1 x and cos −1 𝑥 Step 2 𝑑(sin−1 (𝑥)) 1 = 𝑑𝑥 √1 − 𝑥 2 𝑑(cos −1 (𝑥)) −1 = 𝑑𝑥 √1 − 𝑥 2 Step 3 Therefore, The derivatives of sin−1 𝑥 and cos −1 𝑥 are additive inverse of one another.