Derivatives #6 - Questions and Answers

Question 1 Find the derivative of 𝑦 = sec ⁡ ( 𝑥 ) csc ⁡ ( 𝑥 ) y=sec(x)csc(x). Answer Step 1: Identify the function and differentiate using the product rule. 𝑦 = sec ⁡ ( 𝑥 ) csc ⁡ ( 𝑥 ) y=sec(x)csc(x) Differentiating: 𝑦 ′ = 𝑑 𝑑 𝑥 [ sec ⁡ ( 𝑥 ) ] csc ⁡ ( 𝑥 ) + sec ⁡ ( 𝑥 ) 𝑑 𝑑 𝑥 [ csc ⁡ ( 𝑥 ) ] y ′ = dx d ​ [sec(x)]csc(x)+sec(x) dx d ​ [csc(x)] Step 2: Apply the derivatives of sec ⁡ ( 𝑥 ) sec(x) and csc ⁡ ( 𝑥 ) csc(x): 𝑑 𝑑 𝑥 [ sec ⁡ ( 𝑥 ) ] = sec ⁡ ( 𝑥 ) tan ⁡ ( 𝑥 ) , 𝑑 𝑑 𝑥 [ csc ⁡ ( 𝑥 ) ] = − csc ⁡ ( 𝑥 ) cot ⁡ ( 𝑥 ) dx d ​ [sec(x)]=sec(x)tan(x), dx d ​ [csc(x)]=−csc(x)cot(x) Substitute these values: 𝑦 ′ = sec ⁡ ( 𝑥 ) tan ⁡ ( 𝑥 ) csc ⁡ ( 𝑥 ) + sec ⁡ ( 𝑥 ) ( − csc ⁡ ( 𝑥 ) cot ⁡ ( 𝑥 ) ) y ′ =sec(x)tan(x)csc(x)+sec(x)(−csc(x)cot(x)) Simplify: 𝑦 ′ = sec ⁡ ( 𝑥 ) csc ⁡ ( 𝑥 ) ( tan ⁡ ( 𝑥 ) − cot ⁡ ( 𝑥 ) ) y ′ =sec(x)csc(x)(tan(x)−cot(x)) Question 2 How to find the derivative of a fraction? Answer 1 Use the quotient rule: For a function written as a fraction 𝑢 ( 𝑥 ) 𝑣 ( 𝑥 ) v(x) u(x) ​ , the derivative is: 𝑑 𝑑 𝑥 ( 𝑢 𝑣 ) = 𝑢 ′ 𝑣 − 𝑢 𝑣 ′ 𝑣 2 dx d ​ ( v u ​ )= v 2 u ′ v−uv ′ ​ Here, 𝑢 ′ u ′ and 𝑣 ′ v ′ represent the derivatives of 𝑢 u and 𝑣 v, respectively. Answer 2 The quotient rule simplifies the differentiation of any fraction of two functions or expressions. Question 3 What is the derivative of tan ⁡ ( 2 𝑥 ) tan(2x)? Answer 1 Using the chain rule: The derivative of tan ⁡ ( 𝑥 ) tan(x) is sec ⁡ 2 ( 𝑥 ) sec 2 (x). Let 𝑓 ( 𝑥 ) = tan ⁡ ( 2 𝑥 ) f(x)=tan(2x). Then: 𝑑 𝑑 𝑥 tan ⁡ ( 2 𝑥 ) = sec ⁡ 2 ( 2 𝑥 ) ⋅ 𝑑 𝑑 𝑥 ( 2 𝑥 ) dx d ​ tan(2x)=sec 2 (2x)⋅ dx d ​ (2x) 𝑑 𝑑 𝑥 tan ⁡ ( 2 𝑥 ) = sec ⁡ 2 ( 2 𝑥 ) ⋅ 2 dx d ​ tan(2x)=sec 2 (2x)⋅2 Simplify: 𝑑 𝑑 𝑥 tan ⁡ ( 2 𝑥 ) = 2 sec ⁡ 2 ( 2 𝑥 ) dx d ​ tan(2x)=2sec 2 (2x) Answer 2 Treat tan ⁡ ( 2 𝑥 ) tan(2x) as a composite function 𝑓 ( 𝑔 ( 𝑥 ) ) f(g(x)), where 𝑓 ( 𝑥 ) = tan ⁡ ( 𝑥 ) f(x)=tan(x) and 𝑔 ( 𝑥 ) = 2 𝑥 g(x)=2x. Using the chain rule: 𝑑 𝑑 𝑥 tan ⁡ ( 2 𝑥 ) = sec ⁡ 2 ( 2 𝑥 ) ⋅ 𝑑 𝑑 𝑥 ( 2 𝑥 ) = 2 sec ⁡ 2 ( 2 𝑥 ) dx d ​ tan(2x)=sec 2 (2x)⋅ dx d ​ (2x)=2sec 2 (2x) Question 4 What is the derivative of 𝑦 = arctan ⁡ ( 𝑥 ) y=arctan(x)? Answer 1 The derivative of arctan ⁡ ( 𝑥 ) arctan(x) is: 𝑦 ′ = 1 1 + 𝑥 2 y ′ = 1+x 2 1 ​ This can be derived using implicit differentiation: arctan ⁡ ( 𝑦 ) = 𝑥    ⟹    tan ⁡ ( 𝑦 ) = 𝑥 arctan(y)=x⟹tan(y)=x Differentiate implicitly: sec ⁡ 2 ( 𝑦 ) 𝑑 𝑦 𝑑 𝑥 = 1 sec 2 (y) dx dy ​ =1 Rearranging for 𝑑 𝑦 𝑑 𝑥 dx dy ​ : 𝑑 𝑦 𝑑 𝑥 = 1 sec ⁡ 2 ( 𝑦 ) = 1 1 + tan ⁡ 2 ( 𝑦 ) = 1 1 + 𝑥 2 dx dy ​ = sec 2 (y) 1 ​ = 1+tan 2 (y) 1 ​ = 1+x 2 1 ​ Answer 2 Using the same logic, but rearranging: 𝑑 𝑦 𝑑 𝑥 = 1 1 + 𝑥 2 dx dy ​ = 1+x 2 1 ​ Question 5 Find both first partial derivatives. 𝑧 = sin ⁡ ( 5 𝑥 ) cos ⁡ ( 5 𝑦 ) z=sin(5x)cos(5y). Answer Step 1: Partial derivative with respect to 𝑥 x: Differentiate 𝑧 z partially with respect to 𝑥 x, treating 𝑦 y as a constant: ∂ 𝑧 ∂ 𝑥 = ∂ ∂ 𝑥 [ sin ⁡ ( 5 𝑥 ) cos ⁡ ( 5 𝑦 ) ] ∂x ∂z ​ = ∂x ∂ ​ [sin(5x)cos(5y)] Since cos ⁡ ( 5 𝑦 ) cos(5y) is constant with respect to 𝑥 x: ∂ 𝑧 ∂ 𝑥 = cos ⁡ ( 5 𝑦 ) ⋅ ∂ ∂ 𝑥 [ sin ⁡ ( 5 𝑥 ) ] ∂x ∂z ​ =cos(5y)⋅ ∂x ∂ ​ [sin(5x)] The derivative of sin ⁡ ( 5 𝑥 ) sin(5x) is 5 cos ⁡ ( 5 𝑥 ) 5cos(5x): ∂ 𝑧 ∂ 𝑥 = 5 cos ⁡ ( 5 𝑥 ) cos ⁡ ( 5 𝑦 ) ∂x ∂z ​ =5cos(5x)cos(5y) Step 2: Partial derivative with respect to 𝑦 y: Differentiate 𝑧 z partially with respect to 𝑦 y, treating 𝑥 x as a constant: ∂ 𝑧 ∂ 𝑦 = ∂ ∂ 𝑦 [ sin ⁡ ( 5 𝑥 ) cos ⁡ ( 5 𝑦 ) ] ∂y ∂z ​ = ∂y ∂ ​ [sin(5x)cos(5y)] Since sin ⁡ ( 5 𝑥 ) sin(5x) is constant with respect to 𝑦 y: ∂ 𝑧 ∂ 𝑦 = sin ⁡ ( 5 𝑥 ) ⋅ ∂ ∂ 𝑦 [ cos ⁡ ( 5 𝑦 ) ] ∂y ∂z ​ =sin(5x)⋅ ∂y ∂ ​ [cos(5y)] The derivative of cos ⁡ ( 5 𝑦 ) cos(5y) is − 5 sin ⁡ ( 5 𝑦 ) −5sin(5y): ∂ 𝑧 ∂ 𝑦 = − 5 sin ⁡ ( 5 𝑥 ) sin ⁡ ( 5 𝑦 ) ∂y ∂z ​ =−5sin(5x)sin(5y) Final Results: ∂ 𝑧 ∂ 𝑥 = 5 cos ⁡ ( 5 𝑥 ) cos ⁡ ( 5 𝑦 ) , ∂ 𝑧 ∂ 𝑦 = − 5 sin ⁡ ( 5 𝑥 ) sin ⁡ ( 5 𝑦 ) ∂x ∂z ​ =5cos(5x)cos(5y), ∂y ∂z ​ =−5sin(5x)sin(5y)