Questions and Answers #5 Derivatives

Questions and Answers Sheet #5 Derivatives Question #1 Find all the second partial derivatives. 𝑓(𝑥, 𝑦) = 𝑥 4 𝑦 − 3𝑥 5 𝑦 2 𝑓𝑥𝑥 (𝑥, 𝑦) = 𝑓𝑥𝑦 (𝑥, 𝑦) = 𝑓𝑦𝑥 (𝑥, 𝑦) = 𝑓𝑦𝑦 (𝑥, 𝑦) = Answer: Step 1 Given 𝑓(𝑥, 𝑦) = 𝑥 4 𝑦 − 3𝑥 5 𝑦 2 Step 2 𝑓(𝑥, 𝑦) = 𝑥 4 𝑦 − 3𝑥 5 𝑦 2 𝑓𝑥 (𝑥, 𝑦) = 𝜕 (𝑓(𝑥, 𝑦)) = 4𝑥 3 − 15𝑥 4 𝑦 2 𝜕𝑥 𝑓𝑥𝑦 (𝑥, 𝑦) = 𝜕 𝜕𝑓 ( ) 𝜕𝑦 𝜕𝑥 = 𝜕 (4𝑥 3 − 15𝑥 4 𝑦 2 ) 𝜕𝑦 = 𝜕 𝜕 (4𝑥 3 ) − (15𝑥 4 𝑦 2 ) 𝜕𝑦 𝜕𝑦 = 0 − 15𝑥 4 (2𝑦) 𝑓𝑥𝑦 (𝑥, 𝑦) = −30𝑥 4 𝑦 𝑓𝑥𝑥 = 𝜕 𝜕𝑓 ( ) 𝜕𝑥 𝜕𝑥 = 𝜕 (4𝑥 3 − 15𝑥 4 𝑦 2 ) 𝜕𝑥 = 𝜕 𝜕 (4𝑥 3 ) − (15𝑥 4 𝑦 2 ) 𝜕𝑥 𝜕𝑥 𝑓𝑥𝑥 = 12𝑥 2 − 60𝑥 3 𝑦 2 𝑓𝑦 = = 𝜕 4 (𝑥 𝑦 − 3𝑥 5 𝑦 2 ) 𝜕𝑦 𝜕 4 𝜕 (𝑥 𝑦) − (3𝑥 5 𝑦 2 ) 𝜕𝑦 𝜕𝑦 = 𝑥 4 − 3𝑥 5 (2𝑦) 𝑓𝑦 = 𝑥 4 − 6𝑥 5 𝑦 𝑓𝑦𝑥 = 𝜕 𝜕𝑓 ( ) 𝜕𝑥 𝜕𝑦 = 𝜕 4 (𝑥 − 6𝑥 5 𝑦) 𝜕𝑥 = 𝜕 4 𝜕 (𝑥 ) − (6𝑥 5 𝑦) 𝜕𝑥 𝜕𝑥 𝑓𝑦𝑥 = 3𝑥 3 − 30𝑥 4 𝑦 𝑓𝑦𝑦 = 𝜕 𝜕𝑓 ( ) 𝜕𝑦 𝜕𝑦 = 𝜕 4 (𝑥 − 6𝑥 5 𝑦) 𝜕𝑦 = 𝜕 4 𝜕 (𝑥 ) − (6𝑥 5 𝑦) 𝜕𝑦 𝜕𝑦 = 0 − 6𝑥 5 𝑓𝑦𝑦 = 6𝑥 5 Question #2 Find all of the first order partial derivatives for the following functions: 𝑣 = ln(𝑥 sin 𝑦) + 𝑒 𝑥+𝑦 Answer: Step 1 Derivative, in mathematics, the rate of change of a function with respect to a variable. Derivatives are fundamental to the solution of problems in calculus and differential equation. In algebra, an equation can be defined as a mathematical statement consisting of an equal symbol between two algebraic expression that have the same value. The most basic and common algebraic equations in math consist of one or more variables. Step 2 For the first order partial derivative first of all differentiate the equation partially with respect to x, taking the other variable that is y as constant. 𝜕𝑣 𝜕𝑥 = 𝜕(ln(𝑥 sin(𝑦))+𝑒 𝑥+𝑦 ) 𝜕𝑥 = 1 + (𝑒 𝑥+𝑦 × 1) 𝑥 = 1 + 𝑒 𝑥+𝑦 𝑥 For the first order partial derivative now differentiate the equation partially with respect to y, taking the other variable that is x as constant. 𝜕𝑣 𝜕(ln(𝑥 sin(𝑦)) + 𝑒 𝑥+𝑦 ) = 𝜕𝑥 𝜕𝑥 = 1 + (𝑒 𝑥+𝑦 × 1) 𝑥 = 1 + 𝑒 𝑥+𝑦 𝑥 Question #3 Find an equation for the 𝑛𝑡ℎ derivative of 𝑥 1+𝑥 . Answer: Step 1 The given expression is 𝑥 1+𝑥 . Step 2 Find the 𝑛𝑡ℎ derivative of the function as follows. 𝑦= 𝑥 1+𝑥 𝑦′ = 1 (1 + 𝑥)2 𝑦 ′′ = −2 (1 + 𝑥)3 𝑦 ′′′ = 6 (1 + 𝑥)4 …=⋯ 𝑦 (𝑛) = (1 ∗ 2 ∗ 3 ∗ … ∗ 𝑛)(−1)𝑛+1 (1 + 𝑥)𝑛+1 𝑦 (𝑛) = 𝑛! (−1)𝑛+1 (1 + 𝑥)𝑛+1 Question #4 Find the first partial derivatives of f. 𝑓(𝑟, 𝑠, 𝑡) = 𝑟 2 𝑒 2𝑠 Answer: Step 1 The given function is 𝑓(𝑟, 𝑠, 𝑡) = 𝑟 2 𝑒 2𝑠 Step 2 Now finding the first three partial derivatives by differentiating with respect to one variable keeping the other two constant , we get 𝑓(𝑟, 𝑠, 𝑡) = 𝑟 2 𝑒 2𝑠 𝑐𝑜𝑠 𝜕𝑓 = 2𝑟𝑒 2𝑠 𝜕𝑟 𝜕𝑓 = 2𝑟 2 𝑒 2𝑠 𝜕𝑠 𝜕𝑓 = −𝑟 2 𝑒 2𝑠 𝜕𝑡 Question #5 Find the first partial derivatives. 𝑔(𝑥, 𝑦) = ln(𝑥 6 + 𝑦 6 ) 𝑔𝑥 (𝑥, 𝑦) = 𝑔𝑦 (𝑥, 𝑦) = Answer: Step 1 𝑔(𝑥, 𝑦) = ln(𝑥 6 + 𝑦 6 ) Step 2 𝑑 𝑔𝑥 (𝑥, 𝑦) = 𝑔𝑥 (𝑥, 𝑦) = 𝑔𝑥 (𝑥, 𝑦) = 𝑔𝑦 (𝑥, 𝑦) = = 𝑑𝑥 (ln(𝑥 6 + 𝑦 6 )) 1 ∗ 6𝑥 5 𝑥 6 + 𝑦6 6𝑥 5 + 𝑦6 𝑥6 𝑑 (ln(𝑥 6 + 𝑦 6 )) 𝑑𝑥 1 ∗ 6𝑦 5 𝑥 6 + 𝑦6 𝑔𝑦 (𝑥, 𝑦) = 6𝑦 5 + 𝑦6 𝑥6 𝑔𝑥 (𝑥, 𝑦) = 6𝑥 5 𝑥 6 + 𝑦6 𝑔𝑦 (𝑥, 𝑦) = 6𝑦 5 𝑥 6 + 𝑦6 Question #6 Find y' and y''. 𝑦 = 𝑒𝑒 𝑥 Answer: Step 1 Given expression 𝑦 = 𝑒 𝑒 𝑥 Step 2 Using the chain rule of derivatives If 𝑦 = 𝑓(𝑢), where u is a function of x. Then 𝑑𝑦 𝑑𝑥 = 𝑑𝑦 𝑑𝑢 ∗ 𝑑𝑢 𝑑𝑥 Step 3 Apply the chain rule of derivative in the given expression Let 𝑢 = 𝑒 𝑥 Then 𝑦 = 𝑒 𝑢 Now 𝑦′ = 𝑑𝑦 𝑑𝑥 𝑑𝑦 = 𝑑𝑢 ∗ 𝑑𝑢 𝑑𝑥 𝑢) 𝑑(𝑒 𝑑𝑢 ∗ 𝑑𝑢 𝑑𝑥 𝑦 ′ = (𝑒 𝑢 ) ∗ 𝑑𝑢 𝑑𝑥 ....(1) Step 4 Putting the value of u equation (1) 𝑥 𝑦 ′ = (𝑒 𝑒 ) ∗ 𝑑(𝑒 𝑥 ) 𝑑𝑥 𝑥 𝑦 ′ = (𝑒 𝑒 ) ∗ 𝑒 𝑥 𝑥 𝑦 ′ = 𝑒 𝑥 (𝑒 𝑒 ) … (2) Step 5 For the second derivative Using product rule of derivative If 𝑦 = 𝑢(𝑥)𝑣(𝑥) then 𝑑𝑦 𝑑𝑥 = 𝑢(𝑥)𝑣 ′ (𝑥) + 𝑣(𝑥)𝑢′ (𝑥) Step 6 Apply the product rule of derivatives in equation (2) 𝑥 𝑦 ′′ = 𝑒 𝑥 ∗ 𝑑(𝑒 𝑒 ) 𝑑(𝑒 𝑥 ) 𝑥 + (𝑒 𝑒 ) ∗ 𝑑𝑥 𝑑𝑥 𝑥 𝑥 𝑦 ′′ = 𝑒 𝑥 ∗ (𝑒 𝑥 (𝑒 𝑒 )) + (𝑒 𝑒 ) ∗ 𝑒 𝑥 𝑥 𝑦 ′′ = 𝑒 𝑥 (𝑒 𝑒 )(𝑒 𝑥 + 1) Question #7 Compute the partial derivatives. 𝑧 = ln(8𝑥 2 + 5𝑦 2 ) 𝜕𝑧 = ln(16𝑥 + 5𝑦 2 ) 𝜕𝑥 𝜕𝑧 = 𝜕𝑦 Answer: Step 1 𝑧 = ln(8𝑥 2 + 5𝑦 2 ) 𝜕𝑧 𝜕 = ln(8𝑥 2 + 5𝑦 2 ) 𝜕𝑥 𝜕𝑥 = = 1 ∗ (8 × 2𝑥 + 0) (8𝑥 2 + 5𝑦 2 ) 16𝑥 + 5𝑦 2 ) (8𝑥 2 Step 2 𝑧 = ln(8𝑥 2 + 5𝑦 2 ) 𝜕𝑧 𝜕 = ln(8𝑥 2 + 5𝑦 2 ) 𝜕𝑦 𝜕𝑦 = = (8𝑥 2 1 ∗ (0 + 5 × 2𝑦) + 5𝑦 2 ) 10𝑦 (8𝑥 2 + 5𝑦 2 ) Question #8 Find the derivatives of the function. 𝑓(𝑡) = −2𝑡 2 + 3𝑡 − 6 Answer: To find the derivatives of the following function. 𝑓(𝑡) = −2𝑡 2 + 3𝑡 − 6 We use the following formula 𝑑(𝑥 𝑛 ) = 𝑛𝑥 𝑛−1 𝑑𝑥 𝑑(𝑥) = 1𝑑𝑥 𝑑(𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡) = 0 𝑓(𝑡) = −2𝑡 2 + 3𝑡 − 6 Differentiating f with respect to t, 𝑓 ′ (𝑡) = −2(2)𝑡 2−1 + 3(1) − 0 = −4𝑡 + 3 = 3 − 4𝑡 Question #9 Find the derivatives of the function. 4 ℎ(𝑠) = 𝑠 5 Answer: 4 ℎ(𝑠) = 𝑠 5 4 4 Differentiating h with respect to s, ℎ′ (𝑠) = 𝑠 5−1 5 4 4−5 = 𝑠 5 5 4 −1 = 𝑠5 5 = 4 1 5𝑠 5 = 4 5 5 √𝑠 Question #10 Find the derivatives of the function. ℎ(𝑡) = 𝑡2 𝑡+1 + 2𝑡 + 2 Answer: ℎ(𝑡) = 𝑡+1 𝑡 2 + 2𝑡 + 2 Let 𝑢 = 𝑡 + 1, 𝑣 = 𝑡 2 + 2𝑡 + 2 𝑢′ = 1, 𝑣 = 2𝑡 2−1 + 2(1) + 0 = 2𝑡 + 2 𝑢 ′ We know that, ( ) = 𝑣𝑢′ −𝑢𝑣 ′ 𝑣 𝑣2 Differentiating h with respect to t, ℎ′ (𝑡) = ( = (𝑡 2 + 2𝑡 + 2)(1) − (𝑡 + 1)(2𝑡 + 2) (𝑡 2 + 2𝑡 + 2)2 = 𝑡 2 + 2𝑡 + 2 − (2𝑡 2 + 2𝑡 + 2𝑡 + 2) (𝑡 2 + 2𝑡 + 2)2 = 𝑡 2 + 2𝑡 + 2 − 2𝑡 2 − 4𝑡 − 2 (𝑡 2 + 2𝑡 + 2)2 = = 𝑡+1 𝑡 2 +2𝑡+2 ) ′ −𝑡 2 − 2𝑡 + 2𝑡 + 2)2 (𝑡 2 −𝑡(𝑡 + 2) (𝑡 2 + 2𝑡 + 2)2 Question #11 Find the derivatives. 𝑦 = (𝑥 − 1)(𝑥 + 2)(𝑥 + 3) Answer: Step 1 Given function is 𝑦 = (𝑥 − 1)(𝑥 + 2)(𝑥 + 3). To find the derivatives of all orders of the function. Solution: On simplifying the given function can be written as 𝑦 = (𝑥 − 1)[𝑥(𝑥 + 3) + 2(𝑥 + 3)] = (𝑥 − 1)[𝑥 2 + 3𝑥 + 2𝑥 + 6] = (𝑥 − 1)(𝑥 2 + 5𝑥 + 6) = 𝑥(𝑥 2 + 5𝑥 + 6) − 1(𝑥 2 + 5𝑥 + 6) = 𝑥 3 + 5𝑥 2 + 6𝑥 − 𝑥 2 − 5𝑥 − 6 = 𝑥 3 + 4𝑥 2 + 𝑥 − 6 So, we have 𝑦 = 𝑥 3 + 4𝑥 2 + 𝑥 − 6. Step 2 Power rule of differentiation states that: 𝑑 𝑛 (𝑥 ) = 𝑛𝑥 𝑛−1 𝑑𝑥 Finding first derivative of the function. 𝑦′ = 𝑑 3 (𝑥 + 4𝑥 2 + 𝑥 − 6) 𝑑𝑥 = 3𝑥 2 + 8𝑥 + 1 − 0 = 3𝑥 2 + 8𝑥 + 1 Now finding second derivative of the function. 𝑦 ′′ = 𝑑 𝑑𝑥 (3𝑥 2 + 8𝑥 + 1) = 3 ∗ 2𝑥 + 8 ∗ 1 + 0 = 6𝑥 + 8 Step 3 Now finding third derivative of the function. 𝑦 ′′′ = 𝑑 𝑑𝑥 (6𝑥 + 8) =6∗1+0 =6 We know that derivative of a constant function is zero. So, if we find further derivative then it will be zero. Hence, 𝑦 ′ = 3𝑥 2 + 8𝑥 + 1, 𝑦 ′′ = 6𝑥 + 8 and 𝑦 ′′′ = 6. Question #12 Find the derivatives of the function. 𝑦= 5𝑥−7 2𝑥+1 The derivative is 𝑦 ′ = Answer: Step 1 Consider the function; 𝑦= 5𝑥 − 7 2𝑥 + 1 Step 2 Apply the quotient rules of derivatives, Differentiating with respect to “x” 𝑑 𝑑𝑥 𝑦= 𝑦′ = 𝑑 ( 5𝑥−7 𝑑𝑥 2𝑥+1 (2𝑥 + 1) ) 𝑑 𝑑 (5𝑥 − 7) − (5𝑥 − 7) (2𝑥 + 1) 𝑑𝑥 𝑑𝑥 (2𝑥 + 1)2 = (2𝑥 + 1)(5 ∗ 1 − 0) − (5𝑥 − 7)(2 ∗ 1 + 0) (2𝑥 + 1)2 = 10𝑥 + 5 − (10𝑥 − 14) (2𝑥 + 1)2 = 10𝑥 + 5 − 10𝑥 + 14 (2𝑥 + 1)2 𝑦′ = 19 (2𝑥 + 1)2 Question #13 Find the derivatives of the function defined as follows. 𝑦 = sin 𝑒 10𝑥 Answer: Step 1 Use the chain rule of derivation to find the derivative of 𝑦 = sin(𝑒 10𝑥 ). Find the derivative of sin(𝑥) assuming 𝑒 10𝑥 as x and then substitute 𝑒 10𝑥 for x. 𝑦 = sin(𝑒 10𝑥 ) 𝑑𝑦 𝑑(sin(𝑒 10𝑥 )) = 𝑑𝑥 𝑑𝑥 = cos(𝑒 10𝑥 ) Step 2 Using the chain rule derivate the parameter inside cosine function by first finding derivative of 𝑒 𝑥 assuming 10x as x then find the derivative of 10x with respect to x and multiplying all the values of the derivatives to find the derivative of the function 𝑦 = sin 𝑒 10𝑥 𝑦 = sin(𝑒 10𝑥 ) 𝑑𝑦 𝑑(sin(𝑒 10𝑥 )) = 𝑑𝑥 𝑑𝑥 = cos(𝑒 10𝑥 ) = (cos(𝑒 10𝑥 )) ( 𝑑(𝑒 10𝑥 ) 𝑑(10𝑥) 𝑑𝑥 𝑑𝑥 )( ) = cos(𝑒 10𝑥 ) (𝑒 10𝑥 )(10) = 10𝑒 10𝑥 cos(𝑒 10𝑥 ) Question #14 Use the Power Rule to compute the derivative. 𝑑 0.35 𝑥 𝑑𝑥 Answer: Step 1:Introduction The power rule of derivative is given by: 𝑑 𝑛 (𝑥 ) = 𝑛𝑥 𝑛−1 𝑑𝑥 Step 2:To determine Evaluate: 𝑑 0.35 (𝑥 ) 𝑑𝑥 Step 3:Solution Consider 𝑑 𝑑𝑥 (𝑥 0.35 ) Using power rule of derivatives, we get, 𝑑 𝑑𝑥 (𝑥 0.35 ) = 0.35𝑥 0.35−1 = 0.35𝑥 −0.65 Hence, 𝑑 𝑑𝑥 (𝑥 0.35 ) = 0.35𝑥 −0.65 Step 4:Conclusion Hence, 𝑑 𝑑𝑥 (𝑥 0.35 ) = 0.35𝑥 −0.65 Question #15 The functions 𝑦 = 𝑒 𝑥 does. 3 and 𝑦 = 𝑥 3 𝑒 𝑥 3 do not have elementary anti-derivatives, but 𝑦 = (1 + 3𝑥 3 )𝑒 𝑥 Evaluate 3 ∫(1 + 3𝑥 3 )𝑒 𝑥 𝑑𝑥. Answer: Step 1 3 Given, ∫(1 + 3𝑥 3 )𝑒 𝑥 𝑑𝑥 Step 2 Apply integration by parts, 𝑑 3 3 3 ∫(1 + 3𝑥 3 )𝑒 𝑥 𝑑𝑥 = (1 + 3𝑥 3 ) ∫ 𝑒 𝑥 𝑑𝑥 − ∫ ( (1 + 3𝑥 3 ) ∫ 𝑒 𝑥 𝑑𝑥) 𝑑𝑥 𝑑𝑥 3 = (1 + 3𝑥 3 ) = 3 𝑒𝑥 𝑒𝑥 2) − ∫ ((9𝑥 ) 𝑑𝑥 3𝑥 2 3𝑥 2 1 + 3𝑥 3 𝑥 3 3 𝑒 − 3 ∫ 𝑒 𝑥 𝑑𝑥 3𝑥 2 3 3 = 1 + 3𝑥 3 𝑥 3 𝑒𝑥 𝑒 −3∗ 2 2 3𝑥 3𝑥 = 1 + 3𝑥 3 𝑥 3 𝑒 𝑥 𝑒 − 2 3𝑥 2 𝑥 = 1 + 3𝑥 3 − 3 𝑥 3 𝑒 3𝑥 2 = 3𝑥 3 − 2 𝑥 3 𝑒 3𝑥 2 3 Question #16 Find all the second partial derivatives. 𝑓(𝑥, 𝑦) = 𝑥 6 𝑦 − 2𝑥 3 𝑦 2 𝑓𝑥𝑥 (𝑥, 𝑦) = 𝑓𝑥𝑦 (𝑥, 𝑦) = 𝑓𝑦𝑥 (𝑥, 𝑦) = 𝑓𝑦𝑦 (𝑥, 𝑦) = Answer: Step 1 Given 𝑓(𝑥, 𝑦) = 𝑥 6 𝑦 − 2𝑥 3 𝑦 2 Step 2 𝑓𝑥𝑥 = = 𝜕(𝑓𝑥) 𝜕𝑥 𝜕(6𝑥 5 𝑦 − 6𝑥 2 𝑦 2 ) 𝜕𝑥 = 6𝑦 𝜕(𝑥 5 ) − 6𝑦 2 [2𝑥] 𝜕𝑥 𝑓𝑥𝑥 = 6𝑥𝑦[5𝑥 3 − 2𝑦] 𝑓𝑥𝑦 = 6𝑥 2 (𝑥 3 − 2𝑦) 𝑓𝑦𝑥 = 6𝑥 2 (𝑥 3 − 2𝑦) 𝑓𝑦𝑦 = −4𝑥 3 Question #17 Find the derivative with respect to the independent variable. 𝑓(𝑥) = 9 cos 𝑥 − 2 sin 𝑥 𝑓 ′ (𝑥) = Answer: Step 1 We have to find derivatives of function: 𝑓(𝑥) = 9 cos 𝑥 − 2 sin 𝑥 We know the derivatives of trigonometric ratios: 𝑑 cos 𝑥 = − sin 𝑥 𝑑𝑥 𝑑 sin 𝑥 = cos 𝑥 𝑑𝑥 Step 2 So differentiating the function 𝑓(𝑥) = 9 cos 𝑥 − 2 sin 𝑥 with respect to 'x', we get 𝑓(𝑥) = 9 cos 𝑥 − 2 sin 𝑥 𝑓 ′ (𝑥) = 9 𝑑 cos 𝑥 𝑑 sin 𝑥 −2 𝑑𝑥 𝑑𝑥 = 9(− sin 𝑥) − 2(cos 𝑥) = −9 sin − 2 cos 𝑥 Hence, value of f'(x) is −9 sin − 2 cos 𝑥. Question #18 Find all the second partial derivatives. 𝑓(𝑥, 𝑦) = 𝑥 4 𝑦 − 2𝑥 5 𝑦 2 𝑓𝑥𝑥 (𝑥, 𝑦) = 𝑓𝑥𝑦 (𝑥, 𝑦) = 𝑓𝑦𝑥 (𝑥, 𝑦) = 𝑓𝑦𝑦 (𝑥, 𝑦) = Answer: Given, 𝑓(𝑥, 𝑦) = 𝑥 4 𝑦 − 2𝑥 5 𝑦 2 Partially differentiating with respect to x, we get 𝑓𝑥 (𝑥, 𝑦) = 𝑦 ∗ 4𝑥 3 − 2𝑦 2 ∗ 5𝑥 4 = 4𝑥 3 𝑦 − 10𝑥 4 𝑦 2 Partially differentiating with respect to y, we get 𝑓𝑦 (𝑥, 𝑦) = 𝑥 4 ∗ 1 − 2𝑥 5 ∗ 2𝑦 = 𝑥 4 − 4𝑥 5 𝑦 Step 2 Partially differentiating 𝑓𝑥 (𝑥, 𝑦) = 4𝑥 3 𝑦 − 10𝑥 4 𝑦 2 with respect to x, we get 𝑓𝑥𝑥 (𝑥, 𝑦) = 4𝑦 ∗ 3𝑥 2 − 10𝑦 2 ∗ 4𝑥 3 = 12𝑥 2 𝑦 − 40𝑥 3 𝑦 2 Partially differentiating 𝑓𝑥 (𝑥, 𝑦) = 4𝑥 3 𝑦 − 10𝑥 4 𝑦 2 with respect to y, we get 𝑓𝑥𝑦 (𝑥, 𝑦) = 4𝑥 3 ∗ 1 − 10𝑥 4 ∗ 2𝑦 = 4𝑥 3 − 20𝑥 4 𝑦 Partially differentiating 𝑓𝑦 (𝑥, 𝑦) = 𝑥 4 − 4𝑥 5 𝑦 with respect to x, we get 𝑓𝑦𝑥 (𝑥, 𝑦) = 4𝑥 3 − 4𝑦 ∗ 5𝑥 4 = 4𝑥 3 − 20𝑥 4 𝑦 Partially differentiating 𝑓𝑦 (𝑥, 𝑦) = 𝑥 4 − 4𝑥 5 𝑦 with respect to y, we get 𝑓𝑦𝑦 (𝑥, 𝑦) = 0 − 4𝑥 5 ∗ 1 = −4𝑥 5 Question #19 Find all first partial derivatives. 𝑓(𝑥, 𝑦) = 9𝑥 3 + 4𝑦 − 5 𝑓𝑥 (𝑥, 𝑦) = 𝑓𝑦 (𝑥, 𝑦) = Answer: Step 1 𝑓(𝑥, 𝑦) = 9𝑥 3 + 4𝑦 − 5 𝜕 𝜕 (9𝑥 3 + 4𝑦 − 5) (𝑓(𝑥, 𝑦)) = 𝜕𝑥 𝜕𝑥 𝑓𝑥 (𝑥, 𝑦) = 27𝑥 2 Step 2 𝑓𝑥 (𝑥, 𝑦) = 27𝑥 2 𝜕 𝜕 (9𝑥 3 + 4𝑦 − 5) (𝑓(𝑥, 𝑦)) = 𝜕𝑦 𝜕𝑦 𝑓𝑦 (𝑥, 𝑦) = 4 Question #20 Find the derivative of 𝑓(𝑥) = 𝑥 2 + 3𝑥 at x. That is, find f'(x). Answer: Step 1 We have to find the derivative of f(x) at x: 𝑓(𝑥) = 𝑥 2 + 3𝑥 We will use following formula, for derivatives, 𝑑𝑓(𝑥) = 𝑓 ′ (𝑥) 𝑑𝑥 𝑑𝑎𝑥 𝑛 = 𝑎𝑛𝑥 𝑛−1 𝑑𝑥 𝑑𝑥 =1 𝑑𝑥 Step 2 So finding derivative with respect to 'x', we get 𝑑𝑓(𝑥) 𝑑(𝑥 2 + 3𝑥) = 𝑑𝑥 𝑑𝑥 𝑓 ′ (𝑥) = 𝑑𝑥 2 𝑑𝑥 +3 𝑑𝑥 𝑑𝑥 = 2𝑥 2−1 + 3 × 1 = 2𝑥 + 3 Hence, value of f'(x) is 2𝑥 + 3.