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Math Help - Finding derivatives

  1. #1
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    Finding derivatives

    Sorry - I can't change the Title, I put something random in whilst checking my Latex syntax now I can't change it to something more appropriate could a mod please do it for me? Thanks

    I've copied out the next question and my attempt at an answer, could you please advise as to what I have got wrong;

    In the each of the following parts, you should simplify your answers where it is appropriate to do so.

    a)
    i) Write down the derivative of each of the functions

    f(x)=x^6 and g(x)=e^{4x}

    ii) Hence, by using the Product Rule, differentiate the funtion

    k(x)=x^6 e^{4x}


    b)
    i) Write down the derivative of each of the functions

    f(t)=8-t^3 and g(t)=\cos(3t)

    ii) Hence, by using the Quotient Rule, differentiate the funtion

    k(t)=\frac{8-t^3}{cos(3t)} (-\frac{1}{6}\pi<t<\frac{1}{6}\pi)


    c)
    i) Write down the derivative of the function

    f(x)=5 \ln(2x) (x>0)

    ii) Hence, by using the Product Rule, differentiate the funtion

    k(x)=\sin(5 \ln(2x)) (x>0)
    a)
    i) The derivatives are;

    f(x)=x^6

    f'(x)=6x^5


    g(x)=e^{4x}

    g'(x)=e^{4x}

    ii) Differentiate the function using the product rule

    k(x)=x^6 e^{4x}

    k'(x)=(6x^5)(e^{4x}) + (x^6)(e^{4x})


    b)
    i) The derivatives are;

    f(t)=8-t^3

    f'(t)=-3t^2


    g(t)=\cos(3t)

    g'(t)=-\sin(3t)

    ii)Differentiate the function using the Quotient rule

    k(t)=\frac{8-t^3}{cos(3t)} (-\frac{1}{6}\pi<t<\frac{1}{6}\pi)

    k'(t)\frac{\cos(3t)(-3t^2)-(8-t^3)(\sin(3t))}{\cos(3t)^2}


    c)
    i) The derivatives are;

    f(x)=5 \ln(2x) (x>0)

    f'(x)=\frac{1}{x} 5(2x)

    ii)Differentiate the function using the Composite Rule

    k(x)=\sin(5 \ln(2x)) (x>0)

    k(x)=\sin(5 \ln(2x)) \frac{1}{x} 5(2x)

    Thank you
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  2. #2
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    Quote Originally Posted by RhysGM View Post
    ...
    ii) Differentiate the function using the product rule

    k(x)=x^6 e^{4x}

    k'(x)=(6x^5)(e^{4x}) + (x^6)(e^{4x})= \color{blue}\bold{x^5 \cdot e^{4x}(6+x)}


    ...

    c)
    i) The derivatives are;

    f(x)=5 \ln(2x) (x>0)

    f'(x)=\frac{1}{\bold{\color{red}2}\color{black}x} 5(2) = \dfrac5{x}

    ii)Differentiate the function using the Composite Rule

    k(x)=\sin(5 \ln(2x)) (x>0)

    \bold{\color{blue}k'(x)=\cos(5 \ln(2x)) \frac{5}{x}}

    Thank you
    ...
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  3. #3
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    Derivitives

    Quote Originally Posted by earboth View Post
    k'(x)=(6x^5)(e^{4x}) + (x^6)(e^{4x})= \color{blue}\bold{x^5 \cdot e^{4x}(6+x)}



    Thank you for your corrections. But could you step me through how you simplified the above.
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  4. #4
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    Quote Originally Posted by RhysGM View Post
    Sorry - I can't change the Title, I put something random in whilst checking my Latex syntax now I can't change it to something more appropriate could a mod please do it for me? Thanks

    I've copied out the next question and my attempt at an answer, could you please advise as to what I have got wrong;

    a)
    i) The derivatives are;

    f(x)=x^6

    f'(x)=6x^5


    g(x)=e^{4x}

    g'(x)=e^{4x}
    No. g'(x)= 4e^{4x}
    Chain rule!

    ii) Differentiate the function using the product rule

    k(x)=x^6 e^{4x}

    k'(x)=(6x^5)(e^{4x}) + (x^6)(e^{4x})
    k'(x)= 6x^5(e^{4x})+ x^6(4e^{4x})= (4x^6+ 6x^5)e^{4x}


    b)
    i) The derivatives are;

    f(t)=8-t^3

    f'(t)=-3t^2


    g(t)=\cos(3t)

    g'(t)=-\sin(3t)
    No, g'(t)= -3 sin(3t)
    Chain rule again.

    ii)Differentiate the function using the Quotient rule

    k(t)=\frac{8-t^3}{cos(3t)} (-\frac{1}{6}\pi<t<\frac{1}{6}\pi)

    k'(t)\frac{\cos(3t)(-3t^2)-(8-t^3)(\sin(3t))}{\cos(3t)^2}
     k'(t)= \frac{-3t^2cos(3t)+ 3(8- t^3)sin(3t)}{cos^2(3t)}



    c)
    i) The derivatives are;

    f(x)=5 \ln(2x) (x>0)

    f'(x)=\frac{1}{x} 5(2x)
    No. Notice that your solution is the same as 2/5, a constant.
    f'= \frac{5}{2x}\left(2\right)= \frac{5}{x}
    Notice that the "2" has disappeared here. That is because 5ln(2x)= 5(ln(x)+ ln(2))= 5ln(x)+ 5ln(2), and the derivative of the constant 5ln(x) is 0.

    ii)Differentiate the function using the Composite Rule
    I would say "chain rule"

    k(x)=\sin(5 \ln(2x)) (x>0)

    k(x)=\sin(5 \ln(2x)) \frac{1}{x} 5(2x)
    k'(x)= \frac{5 cos(5ln(2x))}{x}
    The derivative of sin(x) is cos(x), not sin(x).

    Thank you
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  5. #5
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    I think I've figured it out;

     k'(x) = (6x^5)(e^{4x}) + (x^6)(e^{4x})

     k'(x) = 6(x^5)(e^{4x}) + x(x^5)(e^{4x})

     k'(x) = (x^5)(e^{4x})(6+x)
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  6. #6
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    Quote Originally Posted by HallsofIvy View Post
    ...

    k'(x)= 6x^5(e^{4x})+ x^6(4e^{4x})= (4x^6+ 6x^5)e^{4x}

    ...
    Quote Originally Posted by RhysGM View Post
    I think I've figured it out; Unfortunately no. See above HallsofIvy's post

     k'(x) = (6x^5)(e^{4x}) + (x^6)(e^{4x})

    [...
     k'(x) = (6x^5)(e^{4x}) + (x^6)(e^{4x} \cdot 4) = x^5 \cdot e^{4x} \cdot 6+ x^5 \cdot e^{4x} \cdot 4x = x^5 \cdot e^{4x}(6+4x)

    The common factor is x^5 \cdot e^{4x} which I put before the bracket.
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