Derivatives #7 - Questions and Answers

Question 1 Find the first partial derivatives of the function 𝑓 ( 𝑥 , 𝑦 ) = 4 𝑥 2 𝑦 3 + 3 𝑥 2 𝑦 3 + 10 f(x,y)=4x 2 y 3 +3x 2 y 3 +10. Answer Step 1: First partial derivative with respect to 𝑥 x: 𝑓 𝑥 ( 𝑥 , 𝑦 ) = ∂ ∂ 𝑥 ( 4 𝑥 2 𝑦 3 + 3 𝑥 2 𝑦 3 + 10 ) f x ​ (x,y)= ∂x ∂ ​ (4x 2 y 3 +3x 2 y 3 +10) Differentiate term by term: 𝑓 𝑥 ( 𝑥 , 𝑦 ) = ∂ ∂ 𝑥 ( 4 𝑥 2 𝑦 3 ) + ∂ ∂ 𝑥 ( 3 𝑥 2 𝑦 3 ) + ∂ ∂ 𝑥 ( 10 ) f x ​ (x,y)= ∂x ∂ ​ (4x 2 y 3 )+ ∂x ∂ ​ (3x 2 y 3 )+ ∂x ∂ ​ (10) Simplify: 𝑓 𝑥 ( 𝑥 , 𝑦 ) = 8 𝑥 𝑦 3 + 6 𝑥 𝑦 3 + 0 = 12 𝑥 𝑦 3 f x ​ (x,y)=8xy 3 +6xy 3 +0=12xy 3 Step 2: First partial derivative with respect to 𝑦 y: 𝑓 𝑦 ( 𝑥 , 𝑦 ) = ∂ ∂ 𝑦 ( 4 𝑥 2 𝑦 3 + 3 𝑥 2 𝑦 3 + 10 ) f y ​ (x,y)= ∂y ∂ ​ (4x 2 y 3 +3x 2 y 3 +10) Differentiate term by term: 𝑓 𝑦 ( 𝑥 , 𝑦 ) = ∂ ∂ 𝑦 ( 4 𝑥 2 𝑦 3 ) + ∂ ∂ 𝑦 ( 3 𝑥 2 𝑦 3 ) + ∂ ∂ 𝑦 ( 10 ) f y ​ (x,y)= ∂y ∂ ​ (4x 2 y 3 )+ ∂y ∂ ​ (3x 2 y 3 )+ ∂y ∂ ​ (10) Simplify: 𝑓 𝑦 ( 𝑥 , 𝑦 ) = 12 𝑥 2 𝑦 2 + 9 𝑥 2 𝑦 2 + 0 = 21 𝑥 2 𝑦 2 f y ​ (x,y)=12x 2 y 2 +9x 2 y 2 +0=21x 2 y 2 Final Results: 𝑓 𝑥 ( 𝑥 , 𝑦 ) = 12 𝑥 𝑦 3 , 𝑓 𝑦 ( 𝑥 , 𝑦 ) = 21 𝑥 2 𝑦 2 f x ​ (x,y)=12xy 3 ,f y ​ (x,y)=21x 2 y 2 Question 2 Find the first and second derivatives of 𝑦 = 6 𝑥 2 − 10 𝑥 − 5 𝑥 − 2 y=6x 2 −10x−5x −2 . Answer Step 1: First derivative Using the power rule 𝑑 𝑑 𝑥 [ 𝑥 𝑛 ] = 𝑛 𝑥 𝑛 − 1 dx d ​ [x n ]=nx n−1 : 𝑑 𝑦 𝑑 𝑥 = 𝑑 𝑑 𝑥 ( 6 𝑥 2 ) − 𝑑 𝑑 𝑥 ( 10 𝑥 ) − 𝑑 𝑑 𝑥 ( 5 𝑥 − 2 ) dx dy ​ = dx d ​ (6x 2 )− dx d ​ (10x)− dx d ​ (5x −2 ) Simplify: 𝑑 𝑦 𝑑 𝑥 = 12 𝑥 − 10 + 10 𝑥 − 3 dx dy ​ =12x−10+10x −3 Step 2: Second derivative Differentiate 𝑑 𝑦 𝑑 𝑥 = 12 𝑥 − 10 + 10 𝑥 − 3 dx dy ​ =12x−10+10x −3 : 𝑑 2 𝑦 𝑑 𝑥 2 = 𝑑 𝑑 𝑥 ( 12 𝑥 ) − 𝑑 𝑑 𝑥 ( 10 ) + 𝑑 𝑑 𝑥 ( 10 𝑥 − 3 ) dx 2 d 2 y ​ = dx d ​ (12x)− dx d ​ (10)+ dx d ​ (10x −3 ) Simplify: 𝑑 2 𝑦 𝑑 𝑥 2 = 12 + 0 − 30 𝑥 − 4 dx 2 d 2 y ​ =12+0−30x −4 Final Results: 𝑑 𝑦 𝑑 𝑥 = 12 𝑥 − 10 + 10 𝑥 − 3 , 𝑑 2 𝑦 𝑑 𝑥 2 = 12 − 30 𝑥 − 4 dx dy ​ =12x−10+10x −3 , dx 2 d 2 y ​ =12−30x −4 Question 3 Find the derivative of 𝑦 = 4 𝑥 𝑒 − 4 𝑥 + 4 𝑒 𝑥 4 y=4xe −4x +4e x 4 . Answer Step 1: Differentiate term by term using product and chain rules. 𝑑 𝑦 𝑑 𝑥 = 𝑑 𝑑 𝑥 [ 4 𝑥 𝑒 − 4 𝑥 ] + 𝑑 𝑑 𝑥 [ 4 𝑒 𝑥 4 ] dx dy ​ = dx d ​ [4xe −4x ]+ dx d ​ [4e x 4 ] For the first term 4 𝑥 𝑒 − 4 𝑥 4xe −4x , apply the product rule: 𝑑 𝑑 𝑥 [ 4 𝑥 𝑒 − 4 𝑥 ] = 4 ( 𝑑 𝑑 𝑥 [ 𝑥 ] 𝑒 − 4 𝑥 + 𝑥 𝑑 𝑑 𝑥 [ 𝑒 − 4 𝑥 ] ) dx d ​ [4xe −4x ]=4( dx d ​ [x]e −4x +x dx d ​ [e −4x ]) 𝑑 𝑑 𝑥 [ 𝑒 − 4 𝑥 ] = − 4 𝑒 − 4 𝑥 dx d ​ [e −4x ]=−4e −4x Simplify: 𝑑 𝑑 𝑥 [ 4 𝑥 𝑒 − 4 𝑥 ] = 4 ( 𝑒 − 4 𝑥 + 𝑥 ( − 4 𝑒 − 4 𝑥 ) ) = 4 𝑒 − 4 𝑥 − 16 𝑥 𝑒 − 4 𝑥 dx d ​ [4xe −4x ]=4(e −4x +x(−4e −4x ))=4e −4x −16xe −4x For the second term 4 𝑒 𝑥 4 4e x 4 , apply the chain rule: 𝑑 𝑑 𝑥 [ 4 𝑒 𝑥 4 ] = 4 ⋅ 𝑒 𝑥 4 ⋅ 4 𝑥 3 = 16 𝑥 3 𝑒 𝑥 4 dx d ​ [4e x 4 ]=4⋅e x 4 ⋅4x 3 =16x 3 e x 4 Combine results: 𝑑 𝑦 𝑑 𝑥 = 4 𝑒 − 4 𝑥 − 16 𝑥 𝑒 − 4 𝑥 + 16 𝑥 3 𝑒 𝑥 4 dx dy ​ =4e −4x −16xe −4x +16x 3 e x 4 Question 4 Find the second-order partial derivatives of 𝑓 ( 𝑥 , 𝑦 ) = sin ⁡ ( 𝑥 𝑦 ) f(x,y)=sin(xy). Answer Step 1: First-order partial derivatives 𝑓 𝑥 = ∂ ∂ 𝑥 [ sin ⁡ ( 𝑥 𝑦 ) ] = cos ⁡ ( 𝑥 𝑦 ) ⋅ 𝑦 f x ​ = ∂x ∂ ​ [sin(xy)]=cos(xy)⋅y 𝑓 𝑦 = ∂ ∂ 𝑦 [ sin ⁡ ( 𝑥 𝑦 ) ] = cos ⁡ ( 𝑥 𝑦 ) ⋅ 𝑥 f y ​ = ∂y ∂ ​ [sin(xy)]=cos(xy)⋅x Step 2: Second-order partial derivatives 𝑓 𝑥 𝑥 f xx ​ : Differentiate 𝑓 𝑥 f x ​ with respect to 𝑥 x: 𝑓 𝑥 𝑥 = ∂ ∂ 𝑥 [ cos ⁡ ( 𝑥 𝑦 ) ⋅ 𝑦 ] = − sin ⁡ ( 𝑥 𝑦 ) ⋅ 𝑦 2 f xx ​ = ∂x ∂ ​ [cos(xy)⋅y]=−sin(xy)⋅y 2 𝑓 𝑦 𝑦 f yy ​ : Differentiate 𝑓 𝑦 f y ​ with respect to 𝑦 y: 𝑓 𝑦 𝑦 = ∂ ∂ 𝑦 [ cos ⁡ ( 𝑥 𝑦 ) ⋅ 𝑥 ] = − sin ⁡ ( 𝑥 𝑦 ) ⋅ 𝑥 2 f yy ​ = ∂y ∂ ​ [cos(xy)⋅x]=−sin(xy)⋅x 2 𝑓 𝑥 𝑦 f xy ​ : Differentiate 𝑓 𝑥 f x ​ with respect to 𝑦 y: 𝑓 𝑥 𝑦 = ∂ ∂ 𝑦 [ cos ⁡ ( 𝑥 𝑦 ) ⋅ 𝑦 ] = − sin ⁡ ( 𝑥 𝑦 ) ⋅ 𝑥 𝑦 + cos ⁡ ( 𝑥 𝑦 ) f xy ​ = ∂y ∂ ​ [cos(xy)⋅y]=−sin(xy)⋅xy+cos(xy) 𝑓 𝑦 𝑥 f yx ​ : Differentiate 𝑓 𝑦 f y ​ with respect to 𝑥 x: 𝑓 𝑦 𝑥 = ∂ ∂ 𝑥 [ cos ⁡ ( 𝑥 𝑦 ) ⋅ 𝑥 ] = − sin ⁡ ( 𝑥 𝑦 ) ⋅ 𝑥 𝑦 + cos ⁡ ( 𝑥 𝑦 ) f yx ​ = ∂x ∂ ​ [cos(xy)⋅x]=−sin(xy)⋅xy+cos(xy) Final Results: 𝑓 𝑥 𝑥 = − sin ⁡ ( 𝑥 𝑦 ) 𝑦 2 , 𝑓 𝑦 𝑦 = − sin ⁡ ( 𝑥 𝑦 ) 𝑥 2 , 𝑓 𝑥 𝑦 = 𝑓 𝑦 𝑥 = − sin ⁡ ( 𝑥 𝑦 ) 𝑥 𝑦 + cos ⁡ ( 𝑥 𝑦 ) f xx ​ =−sin(xy)y 2 ,f yy ​ =−sin(xy)x 2 ,f xy ​ =f yx ​ =−sin(xy)xy+cos(xy) Question 5 Find the first and second derivatives of 𝑦 = 7 𝑥 5 5 − 2 𝑥 + 9 𝑒 𝑥 y= 5 7x 5 ​ −2x+9e x . Answer Step 1: First derivative 𝑑 𝑦 𝑑 𝑥 = 𝑑 𝑑 𝑥 ( 7 𝑥 5 5 − 2 𝑥 + 9 𝑒 𝑥 ) dx dy ​ = dx d ​ ( 5 7x 5 ​ −2x+9e x ) Differentiate term by term: 𝑑 𝑦 𝑑 𝑥 = 7 5 ⋅ 5 𝑥 4 − 2 + 9 𝑒 𝑥 dx dy ​ = 5 7 ​ ⋅5x 4 −2+9e x Simplify: 𝑑 𝑦 𝑑 𝑥 = 7 𝑥 4 − 2 + 9 𝑒 𝑥 dx dy ​ =7x 4 −2+9e x Step 2: Second derivative Differentiate 𝑑 𝑦 𝑑 𝑥 = 7 𝑥 4 − 2 + 9 𝑒 𝑥 dx dy ​ =7x 4 −2+9e x : 𝑑 2 𝑦 𝑑 𝑥 2 = 𝑑 𝑑 𝑥 [ 7 𝑥 4 − 2 + 9 𝑒 𝑥 ] dx 2 d 2 y ​ = dx d ​ [7x 4 −2+9e x ] Simplify: 𝑑 2 𝑦 𝑑 𝑥 2 = 28 𝑥 3 + 9 𝑒 𝑥 dx 2 d 2 y ​ =28x 3 +9e x Final Results: 𝑑 𝑦 𝑑 𝑥 = 7 𝑥 4 − 2 + 9 𝑒 𝑥 , 𝑑 2 𝑦 𝑑 𝑥 2 = 28 𝑥 3 + 9 𝑒 𝑥 dx dy ​ =7x 4 −2+9e x , dx 2 d 2 y ​ =28x 3 +9e x