Derivatives #15 - Questions and Answers

Question 1 Find the derivative of 𝑓 ( 𝑥 ) = 1 𝑥 ln ⁡ 𝑥 f(x)= xlnx 1 ​ : Solution: Use the quotient rule for differentiation: If 𝑓 ( 𝑥 ) = 𝑢 ( 𝑥 ) 𝑣 ( 𝑥 ) f(x)= v(x) u(x) ​ , then: 𝑓 ′ ( 𝑥 ) = 𝑢 ′ ( 𝑥 ) 𝑣 ( 𝑥 ) − 𝑢 ( 𝑥 ) 𝑣 ′ ( 𝑥 ) 𝑣 ( 𝑥 ) 2 . f ′ (x)= v(x) 2 u ′ (x)v(x)−u(x)v ′ (x) ​ . Here: 𝑢 ( 𝑥 ) = 1 u(x)=1 and 𝑣 ( 𝑥 ) = 𝑥 ln ⁡ 𝑥 v(x)=xlnx. 𝑢 ′ ( 𝑥 ) = 0 u ′ (x)=0 (since the derivative of a constant is zero). To find 𝑣 ′ ( 𝑥 ) v ′ (x), use the product rule: 𝑣 ( 𝑥 ) = 𝑥 ln ⁡ 𝑥 v(x)=xlnx, so: 𝑣 ′ ( 𝑥 ) = 𝑑 𝑑 𝑥 [ 𝑥 ln ⁡ 𝑥 ] = 1 ⋅ ln ⁡ 𝑥 + 𝑥 ⋅ 1 𝑥 = ln ⁡ 𝑥 + 1. v ′ (x)= dx d ​ [xlnx]=1⋅lnx+x⋅ x 1 ​ =lnx+1. Substitute into the quotient rule: 𝑓 ′ ( 𝑥 ) = ( 0 ) ( 𝑥 ln ⁡ 𝑥 ) − ( 1 ) ( ln ⁡ 𝑥 + 1 ) ( 𝑥 ln ⁡ 𝑥 ) 2 . f ′ (x)= (xlnx) 2 (0)(xlnx)−(1)(lnx+1) ​ . Simplify the numerator: 𝑓 ′ ( 𝑥 ) = − ( ln ⁡ 𝑥 + 1 ) ( 𝑥 ln ⁡ 𝑥 ) 2 . f ′ (x)= (xlnx) 2 −(lnx+1) ​ . Final Answer: 𝑓 ′ ( 𝑥 ) = − ( ln ⁡ 𝑥 + 1 ) ( 𝑥 ln ⁡ 𝑥 ) 2 . f ′ (x)= (xlnx) 2 −(lnx+1) ​ . Question 2 Find the derivative of 𝑔 ( 𝑧 ) = 𝑒 5 𝑧 − 1 − 𝑒 5 𝑧 g(z)=e 5z−1 −e 5z : Solution: Differentiate 𝑔 ( 𝑧 ) = 𝑒 5 𝑧 − 1 − 𝑒 5 𝑧 g(z)=e 5z−1 −e 5z term by term. For 𝑒 5 𝑧 − 1 e 5z−1 , use the chain rule: 𝑑 𝑑 𝑧 ( 𝑒 5 𝑧 − 1 ) = 𝑒 5 𝑧 − 1 ⋅ 𝑑 𝑑 𝑧 ( 5 𝑧 − 1 ) = 5 𝑒 5 𝑧 − 1 . dz d ​ (e 5z−1 )=e 5z−1 ⋅ dz d ​ (5z−1)=5e 5z−1 . For 𝑒 5 𝑧 e 5z , also use the chain rule: 𝑑 𝑑 𝑧 ( 𝑒 5 𝑧 ) = 𝑒 5 𝑧 ⋅ 𝑑 𝑑 𝑧 ( 5 𝑧 ) = 5 𝑒 5 𝑧 . dz d ​ (e 5z )=e 5z ⋅ dz d ​ (5z)=5e 5z . Combine results: 𝑔 ′ ( 𝑧 ) = 5 𝑒 5 𝑧 − 1 − 5 𝑒 5 𝑧 . g ′ (z)=5e 5z−1 −5e 5z . Final Answer: 𝑔 ′ ( 𝑧 ) = 5 𝑒 5 𝑧 − 1 − 5 𝑒 5 𝑧 . g ′ (z)=5e 5z−1 −5e 5z . Question 3 Find the derivative of 𝑓 ( 𝑥 ) = 𝑒 ln ⁡ 𝑥 − 𝑒 ln ⁡ ( 𝑥 3 ) f(x)=e lnx −e ln(x 3 ) : Solution: Simplify each term using logarithmic properties: 𝑒 ln ⁡ 𝑥 = 𝑥 e lnx =x (since 𝑒 ln ⁡ 𝑎 = 𝑎 e lna =a). 𝑒 ln ⁡ ( 𝑥 3 ) = 𝑥 3 e ln(x 3 ) =x 3 (since ln ⁡ ( 𝑥 3 ) = 3 ln ⁡ 𝑥 ln(x 3 )=3lnx). So: 𝑓 ( 𝑥 ) = 𝑥 − 𝑥 3 . f(x)=x−x 3 . Differentiate 𝑓 ( 𝑥 ) f(x): 𝑓 ′ ( 𝑥 ) = 𝑑 𝑑 𝑥 ( 𝑥 ) − 𝑑 𝑑 𝑥 ( 𝑥 3 ) . f ′ (x)= dx d ​ (x)− dx d ​ (x 3 ). 𝑑 𝑑 𝑥 ( 𝑥 ) = 1 dx d ​ (x)=1. 𝑑 𝑑 𝑥 ( 𝑥 3 ) = 3 𝑥 2 dx d ​ (x 3 )=3x 2 . Combine results: 𝑓 ′ ( 𝑥 ) = 1 − 3 𝑥 2 . f ′ (x)=1−3x 2 . Final Answer: 𝑓 ′ ( 𝑥 ) = 1 − 3 𝑥 2 . f ′ (x)=1−3x 2 . Question 4 Find the derivative of 𝑥 𝑦 = 𝑒 𝑥 2 xy=e x 2 : Solution: Differentiate implicitly with respect to 𝑥 x: Start with 𝑥 𝑦 = 𝑒 𝑥 2 xy=e x 2 . Use the product rule for 𝑥 𝑦 xy: 𝑑 𝑑 𝑥 ( 𝑥 𝑦 ) = 𝑥 𝑑 𝑦 𝑑 𝑥 + 𝑦 𝑑 𝑑 𝑥 ( 𝑥 ) = 𝑥 𝑑 𝑦 𝑑 𝑥 + 𝑦 . dx d ​ (xy)=x dx dy ​ +y dx d ​ (x)=x dx dy ​ +y. Differentiate 𝑒 𝑥 2 e x 2 using the chain rule: 𝑑 𝑑 𝑥 ( 𝑒 𝑥 2 ) = 𝑒 𝑥 2 ⋅ 𝑑 𝑑 𝑥 ( 𝑥 2 ) = 2 𝑥 𝑒 𝑥 2 . dx d ​ (e x 2 )=e x 2 ⋅ dx d ​ (x 2 )=2xe x 2 . Combine results: 𝑥 𝑑 𝑦 𝑑 𝑥 + 𝑦 = 2 𝑥 𝑒 𝑥 2 . x dx dy ​ +y=2xe x 2 . Solve for 𝑑 𝑦 𝑑 𝑥 dx dy ​ : 𝑥 𝑑 𝑦 𝑑 𝑥 = 2 𝑥 𝑒 𝑥 2 − 𝑦 . x dx dy ​ =2xe x 2 −y. 𝑑 𝑦 𝑑 𝑥 = 2 𝑥 𝑒 𝑥 2 − 𝑦 𝑥 . dx dy ​ = x 2xe x 2 −y ​ . Final Answer: 𝑑 𝑦 𝑑 𝑥 = 2 𝑥 𝑒 𝑥 2 − 𝑦 𝑥 . dx dy ​ = x 2xe x 2 −y ​ . Question 5 Find the derivative 𝑑 𝑑 𝑧 [ arctan ⁡ ( 𝑥 2 + 1 ) ] dz d ​ [arctan(x 2 +1)]: Solution: Let 𝑓 ( 𝑧 ) = arctan ⁡ ( 𝑥 2 + 1 ) f(z)=arctan(x 2 +1). Use the chain rule: The derivative of arctan ⁡ ( 𝑢 ) arctan(u) is 1 1 + 𝑢 2 1+u 2 1 ​ , where 𝑢 = 𝑥 2 + 1 u=x 2 +1. Differentiate 𝑢 = 𝑥 2 + 1 u=x 2 +1: 𝑑 𝑑 𝑥 ( 𝑥 2 + 1 ) = 2 𝑥 . dx d ​ (x 2 +1)=2x. Combine results: 𝑑 𝑑 𝑥 [ arctan ⁡ ( 𝑥 2 + 1 ) ] = 1 1 + ( 𝑥 2 + 1 ) 2 ⋅ 2 𝑥 . dx d ​ [arctan(x 2 +1)]= 1+(x 2 +1) 2 1 ​ ⋅2x. Simplify: 𝑑 𝑑 𝑥 [ arctan ⁡ ( 𝑥 2 + 1 ) ] = 2 𝑥 1 + ( 𝑥 2 + 1 ) 2 . dx d ​ [arctan(x 2 +1)]= 1+(x 2 +1) 2 2x ​ . Final Answer: 2 𝑥 1 + ( 𝑥 2 + 1 ) 2 . 1+(x 2 +1) 2 2x ​ . This matches option (d).