7. use the fundamental theorem of calculus to evaluate the derivatives:\na) \\( \\frac { d } { d x } \\left(…

7. use the fundamental theorem of calculus to evaluate the derivatives:\na) \\( \\frac { d } { d x } \\left( \\int _ { 1 } ^ { x } \\ln \\left( 1 + e ^ { u } \\right) d u \\right) = \\)\nc) \\( \\frac { d } { d x } \\left( \\int _ { 2 } ^ { x ^ { 3 } } \\arctan u d u \\right) = \\)\nb) \\( \\frac { d } { d x } \\left( \\int _ { x } ^ { 1 } \\tan \\left( u ^ { 2 } + 3 \\right) d u \\right) = \\)\nd) \\( \\frac { d } { d x } \\left( \\int _ { \\tan x } ^ { x ^ { 2 } } \\ln \\left( u ^ { 2 } + 5 \\right) d u \\right) = \\)

7. use the fundamental theorem of calculus to evaluate the derivatives:\na) \\( \\frac { d } { d x } \\left( \\int _ { 1 } ^ { x } \\ln \\left( 1 + e ^ { u } \\right) d u \\right) = \\)\nc) \\( \\frac { d } { d x } \\left( \\int _ { 2 } ^ { x ^ { 3 } } \\arctan u d u \\right) = \\)\nb) \\( \\frac { d } { d x } \\left( \\int _ { x } ^ { 1 } \\tan \\left( u ^ { 2 } + 3 \\right) d u \\right) = \\)\nd) \\( \\frac { d } { d x } \\left( \\int _ { \\tan x } ^ { x ^ { 2 } } \\ln \\left( u ^ { 2 } + 5 \\right) d u \\right) = \\)

Answer

Explanation:

Step1: Apply the Fundamental Theorem of Calculus Part 1

If (F(x)=\int_{a}^{x}f(u)du), then (F^\prime(x) = f(x)). For part (a), let (f(u)=\ln(1 + e^{u})), (a = 1). By the Fundamental Theorem of Calculus, (\frac{d}{dx}\left(\int_{1}^{x}\ln(1 + e^{u})du\right)=\ln(1 + e^{x}))

Step2: Use the property (\int_{a}^{b}f(u)du=-\int_{b}^{a}f(u)du) and the Fundamental Theorem of Calculus

For part (b), (\int_{x}^{1}\tan(u^{2}+3)du=-\int_{1}^{x}\tan(u^{2}+3)du). Then (\frac{d}{dx}\left(\int_{x}^{1}\tan(u^{2}+3)du\right)=-\frac{d}{dx}\left(\int_{1}^{x}\tan(u^{2}+3)du\right)=-\tan(x^{2}+3))

Step3: Apply the chain - rule with the Fundamental Theorem of Calculus

If (F(x)=\int_{a}^{g(x)}f(u)du), then (F^\prime(x)=f(g(x))\cdot g^\prime(x)) For part (c), let (g(x)=x^{3}), (f(u)=\arctan(u)). Then (g^\prime(x) = 3x^{2}). By the chain - rule and the Fundamental Theorem of Calculus, (\frac{d}{dx}\left(\int_{2}^{x^{3}}\arctan(u)du\right)=\arctan(x^{3})\cdot3x^{2})

Step4: Apply the chain - rule for a more complex upper and lower limit

If (F(x)=\int_{h(x)}^{g(x)}f(u)du=\int_{h(x)}^{a}f(u)du+\int_{a}^{g(x)}f(u)du=-\int_{a}^{h(x)}f(u)du+\int_{a}^{g(x)}f(u)du), then (F^\prime(x)=f(g(x))\cdot g^\prime(x)-f(h(x))\cdot h^\prime(x)) For part (d), let (g(x)=x^{2}), (h(x)=\tan x), (f(u)=\ln(u^{2}+5)). Then (g^\prime(x) = 2x), (h^\prime(x)=\sec^{2}x). By the chain - rule and the Fundamental Theorem of Calculus, (\frac{d}{dx}\left(\int_{\tan x}^{x^{2}}\ln(u^{2}+5)du\right)=\ln((x^{2})^{2}+5)\cdot2x-\ln((\tan x)^{2}+5)\cdot\sec^{2}x=2x\ln(x^{4}+5)-\sec^{2}x\ln(\tan^{2}x + 5))

Answer:

a) (\ln(1 + e^{x})) b) (-\tan(x^{2}+3)) c) (3x^{2}\arctan(x^{3})) d) (2x\ln(x^{4}+5)-\sec^{2}x\ln(\tan^{2}x + 5))