Derivatives #3 - Questions and Answers

Question 1 Find the point on the line 𝑦 = 4 𝑥 + 3 y=4x+3 that is closest to the origin. (x,y) = ( ) Answer Given line is 𝑦 = 4 𝑥 + 3 y=4x+3. Consider the origin ( 0 , 0 ) (0,0) and the point ( 𝑥 , 𝑦 ) (x,y) where 𝑦 = 4 𝑥 + 3 y=4x+3. The distance 𝐷 D is: 𝐷 = 𝑥 2 + ( 4 𝑥 + 3 ) 2 D= x 2 +(4x+3) 2 ​ Differentiate 𝐷 2 D 2 with respect to 𝑥 x: 𝐷 2 = 𝑥 2 + ( 4 𝑥 + 3 ) 2 D 2 =x 2 +(4x+3) 2 Take the derivative and set it to 0: 𝐷 ′ = 8 𝑥 + 24 2 𝑥 2 + 16 𝑥 2 + 24 𝑥 + 9 = 0 D ′ = 2 x 2 +16x 2 +24x+9 ​ 8x+24 ​ =0 Solving for 𝑥 x: 8 𝑥 + 24 = 0    ⟹    𝑥 = − 12 17 8x+24=0⟹x=− 17 12 ​ Now find 𝑦 y: 𝑦 = 4 𝑥 + 3 = 4 ( − 12 17 ) + 3 = 3 17 y=4x+3=4(− 17 12 ​ )+3= 17 3 ​ Thus, the point is: ( 𝑥 , 𝑦 ) = ( − 12 17 , 3 17 ) (x,y)=(− 17 12 ​ , 17 3 ​ ) Question 2 Find the smallest and largest values and points of inflection of the function: 𝑓 ( 𝑥 , 𝑦 ) = 𝑦 4 + 4 𝑦 2 − 𝑥 2 f(x,y)=y 4 +4y 2 −x 2 Answer Step 1: The given function is 𝑓 ( 𝑥 , 𝑦 ) = 𝑦 4 + 4 𝑦 2 − 𝑥 2 f(x,y)=y 4 +4y 2 −x 2 . To find local minimum, maximum, and saddle points, find the partial derivatives. Step 2: 𝑓 𝑥 = − 2 𝑥 , 𝑓 𝑦 = 4 𝑦 3 + 8 𝑦 f x ​ =−2x,f y ​ =4y 3 +8y Equate them to 0: − 2 𝑥 = 0 , 4 𝑦 3 + 8 𝑦 = 0 −2x=0,4y 3 +8y=0 Solve for 𝑦 y: 4 𝑦 ( 𝑦 2 + 2 ) = 0    ⟹    𝑦 = 0 4y(y 2 +2)=0⟹y=0 So, the function has the critical point ( 0 , 0 ) (0,0). Step 3: Find the second partial derivatives: 𝑓 𝑥 𝑥 = − 2 , 𝑓 𝑦 𝑦 = 12 𝑦 2 + 8 , 𝑓 𝑥 𝑦 = 0 f xx ​ =−2,f yy ​ =12y 2 +8,f xy ​ =0 At ( 0 , 0 ) (0,0): 𝐷 = 𝑓 𝑥 𝑥 𝑓 𝑦 𝑦 − ( 𝑓 𝑥 𝑦 ) 2 = ( − 2 ) ( 8 ) − ( 0 ) 2 = − 16 D=f xx ​ f yy ​ −(f xy ​ ) 2 =(−2)(8)−(0) 2 =−16 Here, 𝐷 D is negative. Hence, ( 0 , 0 ) (0,0) is a saddle point. There are no local maxima or minima. Question 3 What is the derivative of 𝑥 3 x 3 ? Answer 1 Using the power rule: 𝑑 𝑑 𝑥 [ 𝑥 𝑛 ] = 𝑛 𝑥 𝑛 − 1 dx d ​ [x n ]=nx n−1 Here 𝑛 = 3 n=3: 𝑑 𝑑 𝑥 [ 𝑥 3 ] = 3 𝑥 2 dx d ​ [x 3 ]=3x 2 Answer 2 Possible derivation using power rule: 𝑑 𝑑 𝑥 [ 𝑥 3 ] = 3 𝑥 2 dx d ​ [x 3 ]=3x 2 Thus, the derivative is: 3 𝑥 2 3x 2 Question 4 How do you find the slope of the tangent line to a curve at a point? Answer Find the specified input value 𝑥 x, and determine the corresponding output 𝑦 y of the function. Differentiate the function to obtain the derivative. Plug the value of 𝑥 x into the derivative to get the slope. Use the slope 𝑚 m and the point ( 𝑥 , 𝑦 ) (x,y) in the tangent line equation 𝑦 = 𝑚 𝑥 + 𝑏 y=mx+b. Solve for 𝑏 b to obtain the full equation of the tangent line. Question 5 Find the second order derivatives of the function cos ⁡ 3 ( 𝑥 ) cos 3 (x). Answer Step 1: Use the identity: cos ⁡ ( 3 𝑥 ) = 4 cos ⁡ 3 ( 𝑥 ) − 3 cos ⁡ ( 𝑥 ) cos(3x)=4cos 3 (x)−3cos(x) Thus: 4 cos ⁡ 3 ( 𝑥 ) = cos ⁡ ( 3 𝑥 ) + 3 cos ⁡ ( 𝑥 ) 4cos 3 (x)=cos(3x)+3cos(x) cos ⁡ 3 ( 𝑥 ) = 1 4 cos ⁡ ( 3 𝑥 ) + 3 4 cos ⁡ ( 𝑥 ) cos 3 (x)= 4 1 ​ cos(3x)+ 4 3 ​ cos(x) Step 2: Differentiate: 𝑓 ′ ( 𝑥 ) = 𝑑 𝑑 𝑥 [ 1 4 cos ⁡ ( 3 𝑥 ) + 3 4 cos ⁡ ( 𝑥 ) ] f ′ (x)= dx d ​ [ 4 1 ​ cos(3x)+ 4 3 ​ cos(x)] 𝑓 ′ ( 𝑥 ) = 1 4 ( − 3 sin ⁡ ( 3 𝑥 ) ) + 3 4 ( − sin ⁡ ( 𝑥 ) ) f ′ (x)= 4 1 ​ (−3sin(3x))+ 4 3 ​ (−sin(x)) 𝑓 ′ ( 𝑥 ) = − 3 4 sin ⁡ ( 3 𝑥 ) − 3 4 sin ⁡ ( 𝑥 ) f ′ (x)=− 4 3 ​ sin(3x)− 4 3 ​ sin(x) Step 3: Differentiate again: 𝑓 ′ ′ ( 𝑥 ) = 𝑑 𝑑 𝑥 [ − 3 4 sin ⁡ ( 3 𝑥 ) − 3 4 sin ⁡ ( 𝑥 ) ] f ′′ (x)= dx d ​ [− 4 3 ​ sin(3x)− 4 3 ​ sin(x)] 𝑓 ′ ′ ( 𝑥 ) = − 3 4 ( 3 cos ⁡ ( 3 𝑥 ) ) − 3 4 ( cos ⁡ ( 𝑥 ) ) f ′′ (x)=− 4 3 ​ (3cos(3x))− 4 3 ​ (cos(x)) 𝑓 ′ ′ ( 𝑥 ) = − 9 4 cos ⁡ ( 3 𝑥 ) − 3 4 cos ⁡ ( 𝑥 ) f ′′ (x)=− 4 9 ​ cos(3x)− 4 3 ​ cos(x) Thus, the second derivative is: 𝑓 ′ ′ ( 𝑥 ) = − 9 4 cos ⁡ ( 3 𝑥 ) − 3 4 cos ⁡ ( 𝑥 ) f ′′ (x)=− 4 9 ​ cos(3x)− 4 3 ​ cos(x)