Derivative of the integral $e^t/t $ from $sqrt{x}$ to $x$












2














I have an integral $frac{e^t}{t}$ from $sqrt{x}$ to x and I have to find the derivative of this. This is for a calc II class so we haven't gone over what Ei(x) is, I'm an sure that this question is an application of FTC pt. 2.



Attempt at the solution:




  1. $$int_{sqrt{x}}^x frac{e^t}{t} dt$$

  2. $$int_{sqrt{x}}^a frac{e^t}{t} dt + int_{a}^x frac{e^t}{t} dt$$ for $sqrt{x}<a<x$

  3. $$int_{a}^x frac{e^t}{t} dt - int_a^{sqrt{x}} frac{e^t}{t} dt$$

  4. Differentiate both sides - $$ frac{e^x}{x} - frac{e^sqrt{x}}{sqrt{x}}$$


The answer in the textbook however is $$frac{2e^x-e^sqrt{x}}{2x}$$



I have tried rearranging by equation by multiplying the second term by $frac{sqrt{x}}{sqrt{x}}$ however my answer does not equal the textbook's answer. I have also graphed both my answer and the textbook's and it is clear they are not equal, due to this I am sure I made an error however I am unsure where.



Thanks for any help you can offer!






calculus







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  • 2




    You need the Chain rule when differentiating the second integral.
    – David Mitra
    Dec 5 '14 at 18:02


















2














I have an integral $frac{e^t}{t}$ from $sqrt{x}$ to x and I have to find the derivative of this. This is for a calc II class so we haven't gone over what Ei(x) is, I'm an sure that this question is an application of FTC pt. 2.



Attempt at the solution:




  1. $$int_{sqrt{x}}^x frac{e^t}{t} dt$$

  2. $$int_{sqrt{x}}^a frac{e^t}{t} dt + int_{a}^x frac{e^t}{t} dt$$ for $sqrt{x}<a<x$

  3. $$int_{a}^x frac{e^t}{t} dt - int_a^{sqrt{x}} frac{e^t}{t} dt$$

  4. Differentiate both sides - $$ frac{e^x}{x} - frac{e^sqrt{x}}{sqrt{x}}$$


The answer in the textbook however is $$frac{2e^x-e^sqrt{x}}{2x}$$



I have tried rearranging by equation by multiplying the second term by $frac{sqrt{x}}{sqrt{x}}$ however my answer does not equal the textbook's answer. I have also graphed both my answer and the textbook's and it is clear they are not equal, due to this I am sure I made an error however I am unsure where.



Thanks for any help you can offer!






calculus







share|cite|improve this question




















  • 2




    You need the Chain rule when differentiating the second integral.
    – David Mitra
    Dec 5 '14 at 18:02
















2












2








2







I have an integral $frac{e^t}{t}$ from $sqrt{x}$ to x and I have to find the derivative of this. This is for a calc II class so we haven't gone over what Ei(x) is, I'm an sure that this question is an application of FTC pt. 2.



Attempt at the solution:




  1. $$int_{sqrt{x}}^x frac{e^t}{t} dt$$

  2. $$int_{sqrt{x}}^a frac{e^t}{t} dt + int_{a}^x frac{e^t}{t} dt$$ for $sqrt{x}<a<x$

  3. $$int_{a}^x frac{e^t}{t} dt - int_a^{sqrt{x}} frac{e^t}{t} dt$$

  4. Differentiate both sides - $$ frac{e^x}{x} - frac{e^sqrt{x}}{sqrt{x}}$$


The answer in the textbook however is $$frac{2e^x-e^sqrt{x}}{2x}$$



I have tried rearranging by equation by multiplying the second term by $frac{sqrt{x}}{sqrt{x}}$ however my answer does not equal the textbook's answer. I have also graphed both my answer and the textbook's and it is clear they are not equal, due to this I am sure I made an error however I am unsure where.



Thanks for any help you can offer!






calculus







share|cite|improve this question















I have an integral $frac{e^t}{t}$ from $sqrt{x}$ to x and I have to find the derivative of this. This is for a calc II class so we haven't gone over what Ei(x) is, I'm an sure that this question is an application of FTC pt. 2.



Attempt at the solution:




  1. $$int_{sqrt{x}}^x frac{e^t}{t} dt$$

  2. $$int_{sqrt{x}}^a frac{e^t}{t} dt + int_{a}^x frac{e^t}{t} dt$$ for $sqrt{x}<a<x$

  3. $$int_{a}^x frac{e^t}{t} dt - int_a^{sqrt{x}} frac{e^t}{t} dt$$

  4. Differentiate both sides - $$ frac{e^x}{x} - frac{e^sqrt{x}}{sqrt{x}}$$


The answer in the textbook however is $$frac{2e^x-e^sqrt{x}}{2x}$$



I have tried rearranging by equation by multiplying the second term by $frac{sqrt{x}}{sqrt{x}}$ however my answer does not equal the textbook's answer. I have also graphed both my answer and the textbook's and it is clear they are not equal, due to this I am sure I made an error however I am unsure where.



Thanks for any help you can offer!






calculus




calculus






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edited Dec 5 '14 at 18:14









Joel

14.3k12240




14.3k12240










asked Dec 5 '14 at 18:00









Tyler NTyler N

184




184








  • 2




    You need the Chain rule when differentiating the second integral.
    – David Mitra
    Dec 5 '14 at 18:02
















  • 2




    You need the Chain rule when differentiating the second integral.
    – David Mitra
    Dec 5 '14 at 18:02










2




2




You need the Chain rule when differentiating the second integral.
– David Mitra
Dec 5 '14 at 18:02






You need the Chain rule when differentiating the second integral.
– David Mitra
Dec 5 '14 at 18:02












2 Answers
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Let
$$F(x)=int_a^x frac{e^t}{t}dt$$
Then
$$F(sqrt{x})=int_a^{sqrt{x}} frac{e^t}{t}dt$$
We have
$$frac{d}{dx}(F(sqrt{x}))=frac{F'(sqrt{x})}{2sqrt{x}}$$
by the chain rule, and this is equal to
$$frac{e^{sqrt{x}}}{2x}$$






share|cite|improve this answer





























    0














    You should use Fundamental Theorem of Calculus. By this Theorem, $$frac{d}{dx}int_{sqrt{x}}^x frac{e^t}{t}dt=frac{e^x}{x}frac{d}{dx}(x)-frac{e^sqrt{x}}{sqrt{x}} frac{d}{dx}(sqrt{x}),$$ So we have
    $$frac{d}{dx}int_{sqrt{x}}^x frac{e^t}{t}dt=frac{e^x}{x}-frac{e^sqrt{x}}{sqrt{x}}frac{1}{2sqrt{x}}=frac{2e^x-e^sqrt{x}}{2x}.$$






    share|cite|improve this answer























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      2 Answers
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      2 Answers
      2






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      2














      Let
      $$F(x)=int_a^x frac{e^t}{t}dt$$
      Then
      $$F(sqrt{x})=int_a^{sqrt{x}} frac{e^t}{t}dt$$
      We have
      $$frac{d}{dx}(F(sqrt{x}))=frac{F'(sqrt{x})}{2sqrt{x}}$$
      by the chain rule, and this is equal to
      $$frac{e^{sqrt{x}}}{2x}$$






      share|cite|improve this answer


























        2














        Let
        $$F(x)=int_a^x frac{e^t}{t}dt$$
        Then
        $$F(sqrt{x})=int_a^{sqrt{x}} frac{e^t}{t}dt$$
        We have
        $$frac{d}{dx}(F(sqrt{x}))=frac{F'(sqrt{x})}{2sqrt{x}}$$
        by the chain rule, and this is equal to
        $$frac{e^{sqrt{x}}}{2x}$$






        share|cite|improve this answer
























          2












          2








          2






          Let
          $$F(x)=int_a^x frac{e^t}{t}dt$$
          Then
          $$F(sqrt{x})=int_a^{sqrt{x}} frac{e^t}{t}dt$$
          We have
          $$frac{d}{dx}(F(sqrt{x}))=frac{F'(sqrt{x})}{2sqrt{x}}$$
          by the chain rule, and this is equal to
          $$frac{e^{sqrt{x}}}{2x}$$






          share|cite|improve this answer












          Let
          $$F(x)=int_a^x frac{e^t}{t}dt$$
          Then
          $$F(sqrt{x})=int_a^{sqrt{x}} frac{e^t}{t}dt$$
          We have
          $$frac{d}{dx}(F(sqrt{x}))=frac{F'(sqrt{x})}{2sqrt{x}}$$
          by the chain rule, and this is equal to
          $$frac{e^{sqrt{x}}}{2x}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 5 '14 at 18:49









          Matt SamuelMatt Samuel

          37.4k63665




          37.4k63665























              0














              You should use Fundamental Theorem of Calculus. By this Theorem, $$frac{d}{dx}int_{sqrt{x}}^x frac{e^t}{t}dt=frac{e^x}{x}frac{d}{dx}(x)-frac{e^sqrt{x}}{sqrt{x}} frac{d}{dx}(sqrt{x}),$$ So we have
              $$frac{d}{dx}int_{sqrt{x}}^x frac{e^t}{t}dt=frac{e^x}{x}-frac{e^sqrt{x}}{sqrt{x}}frac{1}{2sqrt{x}}=frac{2e^x-e^sqrt{x}}{2x}.$$






              share|cite|improve this answer




























                0














                You should use Fundamental Theorem of Calculus. By this Theorem, $$frac{d}{dx}int_{sqrt{x}}^x frac{e^t}{t}dt=frac{e^x}{x}frac{d}{dx}(x)-frac{e^sqrt{x}}{sqrt{x}} frac{d}{dx}(sqrt{x}),$$ So we have
                $$frac{d}{dx}int_{sqrt{x}}^x frac{e^t}{t}dt=frac{e^x}{x}-frac{e^sqrt{x}}{sqrt{x}}frac{1}{2sqrt{x}}=frac{2e^x-e^sqrt{x}}{2x}.$$






                share|cite|improve this answer


























                  0












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                  0






                  You should use Fundamental Theorem of Calculus. By this Theorem, $$frac{d}{dx}int_{sqrt{x}}^x frac{e^t}{t}dt=frac{e^x}{x}frac{d}{dx}(x)-frac{e^sqrt{x}}{sqrt{x}} frac{d}{dx}(sqrt{x}),$$ So we have
                  $$frac{d}{dx}int_{sqrt{x}}^x frac{e^t}{t}dt=frac{e^x}{x}-frac{e^sqrt{x}}{sqrt{x}}frac{1}{2sqrt{x}}=frac{2e^x-e^sqrt{x}}{2x}.$$






                  share|cite|improve this answer














                  You should use Fundamental Theorem of Calculus. By this Theorem, $$frac{d}{dx}int_{sqrt{x}}^x frac{e^t}{t}dt=frac{e^x}{x}frac{d}{dx}(x)-frac{e^sqrt{x}}{sqrt{x}} frac{d}{dx}(sqrt{x}),$$ So we have
                  $$frac{d}{dx}int_{sqrt{x}}^x frac{e^t}{t}dt=frac{e^x}{x}-frac{e^sqrt{x}}{sqrt{x}}frac{1}{2sqrt{x}}=frac{2e^x-e^sqrt{x}}{2x}.$$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 6 '14 at 20:49

























                  answered Dec 6 '14 at 20:13









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