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Thread: 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

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

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

    $\displaystyle k(x)=x^6 e^{4x}$


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

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

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

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


    c)
    i) Write down the derivative of the function

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

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

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

    $\displaystyle f(x)=x^6$

    $\displaystyle f'(x)=6x^5$


    $\displaystyle g(x)=e^{4x}$

    $\displaystyle g'(x)=e^{4x}$

    ii) Differentiate the function using the product rule

    $\displaystyle k(x)=x^6 e^{4x}$

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


    b)
    i) The derivatives are;

    $\displaystyle f(t)=8-t^3$

    $\displaystyle f'(t)=-3t^2$


    $\displaystyle g(t)=\cos(3t)$

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

    ii)Differentiate the function using the Quotient rule

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

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


    c)
    i) The derivatives are;

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

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

    ii)Differentiate the function using the Composite Rule

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

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

    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

    $\displaystyle k(x)=x^6 e^{4x}$

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


    ...

    c)
    i) The derivatives are;

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

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

    ii)Differentiate the function using the Composite Rule

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

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



    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;

    $\displaystyle f(x)=x^6$

    $\displaystyle f'(x)=6x^5$


    $\displaystyle g(x)=e^{4x}$

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

    ii) Differentiate the function using the product rule

    $\displaystyle k(x)=x^6 e^{4x}$

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


    b)
    i) The derivatives are;

    $\displaystyle f(t)=8-t^3$

    $\displaystyle f'(t)=-3t^2$


    $\displaystyle g(t)=\cos(3t)$

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

    ii)Differentiate the function using the Quotient rule

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

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



    c)
    i) The derivatives are;

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

    $\displaystyle f'(x)=\frac{1}{x} 5(2x)$
    No. Notice that your solution is the same as 2/5, a constant.
    $\displaystyle f'= \frac{5}{2x}\left(2\right)= \frac{5}{x}$
    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"

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

    $\displaystyle k(x)=\sin(5 \ln(2x)) \frac{1}{x} 5(2x)$
    $\displaystyle k'(x)= \frac{5 cos(5ln(2x))}{x}$
    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;

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

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

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

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

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

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

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

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