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Math Help - Differentiation

  1. #1
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    Differentiation

    Can someone help me revise how to do these...

    x^4cos(x^2)

    tan(x)/1+x^3 - quotient rule

    sin(exp(1+x^2)) - chain rule?
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  2. #2
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    Quote Originally Posted by dankelly07 View Post
    Can someone help me revise how to do these...

    x^4cos(x^2)

    tan(x)/1+x^3 - quotient rule

    sin(exp(1+x^2)) - chain rule?
    1) x^4\cos x^2

    Product Rule: (x^4\cos x^2)' = (x^4)'\cos x^2 + x^4(\cos x^2)'

    Apply chain rule on \cos x^2

     = 4x^3\cos x^2+ x^4 (2x\cos x^2)

     = 4x^3\cos x^2 + 2x^5\cos x^2

    2) \frac{\tan x}{1+x^3}

    Yes try quotient rule, and tell us what you get.

    3) z = \sin e^{1+x^2}

    Yes chain rule

    First sub u = 1+x^2, y = e^u, then z = \sin e^{1+x^2} = \sin y

    \frac{dz}{dx} = \frac{dz}{dy}\frac{dy}{du}\frac{du}{dx}

    So \frac{dz}{dy} = \cos y, \frac{dy}{du} = e^u, \frac{du}{dx} = 2x,

    \frac{dz}{dx} = \frac{dz}{dy}\frac{dy}{du}\frac{du}{dx} = (\cos y)(e^u)(2x) = 2x e^{1+x^2} \sin e^{1+x^2}
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  3. #3
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    Quote Originally Posted by Isomorphism View Post
    1) x^4\cos x^2

    Product Rule: (x^4\cos x^2)' = (x^4)'\cos x^2 + x^4(\cos x^2)'

    Apply chain rule on \cos x^2

     = 4x^3\cos x^2+ x^4 (2x\cos x^2)

     = 4x^3\cos x^2 + 2x^5\cos x^2
    Iso, you forgot to derive cosine itself...

    (x^4\cos x^2)' = (x^4)'\cos x^2 + x^4(\cos x^2)'

     = 4x^3\cos x^2+- x^4 (2x\sin x^2)

     = 4x^3\cos x^2+- 2x^5\sin x^2)
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  4. #4
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    <br />
\begin{gathered}<br />
  u = \tan (x) \hfill \\<br />
  v = 1 + x^3  \hfill \\<br />
   \hfill \\<br />
  \frac{{du}}<br />
{{dx}} = \sec ^2 (x) \hfill \\<br />
  \frac{{dv}}<br />
{{dx}} = 3x^2  \hfill \\ <br />
\end{gathered} <br />
    <br /> <br />
\begin{gathered}<br />
  so \hfill \\<br />
  \frac{{3x^2 \tan (x) - 1 + x^3 \sec ^2 (x)}}<br />
{{(1 + x^3 )^2 }} ? \hfill \\<br />
   \hfill \\ <br />
\end{gathered} <br />
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  5. #5
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    Quote Originally Posted by dankelly07 View Post
    <br />
\begin{gathered}<br />
  u = \tan (x) \hfill \\<br />
  v = 1 + x^3  \hfill \\<br />
   \hfill \\<br />
  \frac{{du}}<br />
{{dx}} = \sec ^2 (x) \hfill \\<br />
  \frac{{dv}}<br />
{{dx}} = 3x^2  \hfill \\ <br />
\end{gathered} <br />
    <br /> <br />
\begin{gathered}<br />
  so \hfill \\<br />
  \frac{{3x^2 \tan (x) - 1 + x^3 \sec ^2 (x)}}<br />
{{(1 + x^3 )^2 }} ? \hfill \\<br />
   \hfill \\ <br />
\end{gathered} <br />
    No. Two problems: first, the quotient rule is \left(\frac uv\right)' = \frac{vu' - uv'}{v^2}; your numerator is backwards. Second, you did not properly distribute the \sec^2x over the  1 + x^3 factor (or did you forget to type the parentheses?)

    You should get:

    \frac d{dx}\left[\frac{\tan x}{1+x^3}\right]=\frac{\left(1 + x^3\right)\sec^2x - 3x^2\tan x}{\left(1 + x^3\right)^2}
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  6. #6
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    hi just looking over this again, could someone just check that example three is correct?



    , then



    So ,

    I get it up to here..

    but below where does sin come from in the last part? should it not be cose^1+x^2...


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  7. #7
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    Quote Originally Posted by dankelly07 View Post
    hi just looking over this again, could someone just check that example three is correct?



    , then



    So ,

    I get it up to here..

    but below where does sin come from in the last part? should it not be cose^1+x^2...


    Yes, it should be 2xe^{1+x^2}\cos (e^{1+x^2}). See below:
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    z=\sin e^{1+x^2}

    \frac{\mathrm{d}z}{\mathrm{d}x} = \frac{\mathrm{d}z}{\mathrm{d}u}.\frac{\mathrm{d}u}  {\mathrm{d}x}

    Let u=e^{1+x^2} \therefore z=\sin u.

    \frac{\mathrm{d}z}{\mathrm{d}x} = \frac{\mathrm{d}(\sin u)}{\mathrm{d}u}.\frac{\mathrm{d}(e^{1+x^2})}{\mat  hrm{d}x}

    =\cos u . \frac{\mathrm{d}(e^{1+x^2})}{\mathrm{d}x} = \cos (e^{1+x^2}) . \frac{\mathrm{d}(e^{1+x^2})}{\mathrm{d}x}

    For \frac{\mathrm{d}(e^{1+x^2})}{\mathrm{d}x}, use chain rule with v=1+x^2

    \therefore \frac{\mathrm{d}(e^{1+x^2})}{\mathrm{d}x} = 2xe^v = 2xe^{1+x^2}

    \frac{\mathrm{d}z}{\mathrm{d}x} = 2xe^{1+x^2}\cos (e^{1+x^2})
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  8. #8
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    hey ppl ..... just joined this forum a lil while ago and i'm impressed by what i've seen so far. although i havent done calculus since 5 years ago (so i'm really really rusty! ) but just looking at the recent posts brings back some wonderful memories.

    amm .... i'm new to all of this but can someone tell me how to get the different mathematical symbols when posting? i mean like how to get the dy/dx or integral sign, exponent, fractions etc. is there a clipboard on the forum with the symbols on it or are they from an external program or what?

    thanks
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  9. #9
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    Quote Originally Posted by ibnashraf View Post
    hey ppl ..... just joined this forum a lil while ago and i'm impressed by what i've seen so far. although i havent done calculus since 5 years ago (so i'm really really rusty! ) but just looking at the recent posts brings back some wonderful memories.

    amm .... i'm new to all of this but can someone tell me how to get the different mathematical symbols when posting? i mean like how to get the dy/dx or integral sign, exponent, fractions etc. is there a clipboard on the forum with the symbols on it or are they from an external program or what?

    thanks
    It's done using LaTeX

    In Square brackets write math, write a mathematical expression using LaTeX code, then write /math in square brackets.

    For help check the LaTeX sub forum...
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