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(1) Derive the derivative of the following functions \[ \begin{aligned} f(x) & ...
May 19, 2024
Solution by Steps
step 1
To find the derivative of f(x) f(x) , we apply the power rule to each term
step 2
The derivative of x \sqrt{x} is 12x−12 \frac{1}{2} x^{-\frac{1}{2}}
step 3
The derivative of 6x3 6 \sqrt[3]{x} is 6⋅13x−23=2x−23 6 \cdot \frac{1}{3} x^{-\frac{2}{3}} = 2 x^{-\frac{2}{3}}
step 4
The derivative of 2x4 2 \sqrt[4]{x} is 2⋅14x−34=12x−34 2 \cdot \frac{1}{4} x^{-\frac{3}{4}} = \frac{1}{2} x^{-\frac{3}{4}}
step 5
Combining these results, we get f′(x)=12x−12+2x−23+12x−34 f'(x) = \frac{1}{2} x^{-\frac{1}{2}} + 2 x^{-\frac{2}{3}} + \frac{1}{2} x^{-\frac{3}{4}}
# For g(x)=3x3−5x+9x g(x) = \frac{3x^3 - 5x + 9}{x} :
step 1
Simplify g(x) g(x) to g(x)=3x2−5+9x g(x) = 3x^2 - 5 + \frac{9}{x}
step 2
The derivative of 3x2 3x^2 is 6x 6x
step 3
The derivative of −5 -5 is 0 0
step 4
The derivative of 9x \frac{9}{x} is −9x2 -\frac{9}{x^2}
step 5
Combining these results, we get g′(x)=6x−9x2 g'(x) = 6x - \frac{9}{x^2}
# For h(x)=(x+2)(x+3)2 h(x) = (x+2)(x+3)^2 :
step 1
Use the product rule: h′(x)=f′(x)g(x)+f(x)g′(x) h'(x) = f'(x)g(x) + f(x)g'(x) , where f(x)=x+2 f(x) = x+2 and g(x)=(x+3)2 g(x) = (x+3)^2
step 2
The derivative of f(x)=x+2 f(x) = x+2 is f′(x)=1 f'(x) = 1
step 3
The derivative of g(x)=(x+3)2 g(x) = (x+3)^2 is g′(x)=2(x+3) g'(x) = 2(x+3)
step 4
Applying the product rule: h′(x)=1⋅(x+3)2+(x+2)⋅2(x+3) h'(x) = 1 \cdot (x+3)^2 + (x+2) \cdot 2(x+3)
step 5
Simplify: h′(x)=(x+3)2+2(x+2)(x+3)=(x+3)[(x+3)+2(x+2)]=(x+3)(3x+7) h'(x) = (x+3)^2 + 2(x+2)(x+3) = (x+3)[(x+3) + 2(x+2)] = (x+3)(3x+7)
Integrals # For A=∫0ln⁡8(ex−e−x) dx A = \int_{0}^{\ln 8} (e^x - e^{-x}) \, dx :
step 1
Integrate each term separately: ∫ex dx=ex \int e^x \, dx = e^x and ∫−e−x dx=e−x \int -e^{-x} \, dx = e^{-x}
step 2
Combine the results: ∫(ex−e−x) dx=ex+e−x \int (e^x - e^{-x}) \, dx = e^x + e^{-x}
step 3
Evaluate from 0 0 to ln⁡8 \ln 8 : [ex+e−x]0ln⁡8 \left[ e^x + e^{-x} \right]_{0}^{\ln 8}
step 4
Substitute the limits: (eln⁡8+e−ln⁡8)−(e0+e0)=8+18−2 (e^{\ln 8} + e^{-\ln 8}) - (e^0 + e^0) = 8 + \frac{1}{8} - 2
step 5
Simplify: 8+18−2=6+18=488+18=498 8 + \frac{1}{8} - 2 = 6 + \frac{1}{8} = \frac{48}{8} + \frac{1}{8} = \frac{49}{8}
# For B=∫01x2(1+2x3)5 dx B = \int_{0}^{1} x^2 (1 + 2x^3)^5 \, dx :
step 1
Use substitution: let u=1+2x3 u = 1 + 2x^3 , then du=6x2 dx du = 6x^2 \, dx
step 2
Rewrite the integral: B=∫01x2(1+2x3)5 dx=∫1316u5 du B = \int_{0}^{1} x^2 (1 + 2x^3)^5 \, dx = \int_{1}^{3} \frac{1}{6} u^5 \, du
step 3
Integrate: ∫u5 du=u66 \int u^5 \, du = \frac{u^6}{6}
step 4
Evaluate from 1 1 to 3 3 : 16[u66]13=136[36−16] \frac{1}{6} \left[ \frac{u^6}{6} \right]_{1}^{3} = \frac{1}{36} \left[ 3^6 - 1^6 \right]
step 5
Simplify: 136(729−1)=136⋅728=72836=1829 \frac{1}{36} (729 - 1) = \frac{1}{36} \cdot 728 = \frac{728}{36} = \frac{182}{9}
Answer
Derivatives: f′(x)=12x−12+2x−23+12x−34 f'(x) = \frac{1}{2} x^{-\frac{1}{2}} + 2 x^{-\frac{2}{3}} + \frac{1}{2} x^{-\frac{3}{4}} , g′(x)=6x−9x2 g'(x) = 6x - \frac{9}{x^2} , h′(x)=(x+3)(3x+7) h'(x) = (x+3)(3x+7)
Integrals: A=498 A = \frac{49}{8} , B=1829 B = \frac{182}{9}
Key Concept
Derivatives and Integrals
Explanation
The derivatives were found using the power rule and product rule, while the integrals were calculated using basic integration techniques and substitution.
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