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Math Help - given f(x) and dy/dx with constant * f(x). K is?

  1. #1
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    given f(x) and dy/dx with constant * f(x). K is?

    How do I go about tackling the problem below?
    <br />
{\rm If }y = 4^x {\rm  then }\frac{{{\rm dy}}}{{{\rm dx}}} = k.4^x {\rm  where k is: ?}<br />

    Is started with trying first principles and came up with this
    <br />
\mathop {{\rm lim}}\limits_{{\rm h} \to {\rm 0}} \frac{{4^{a + h}  - 4^a }}{h}<br />

    I'm stuck
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Craka View Post
    How do I go about tackling the problem below?
    <br />
{\rm If }y = 4^x {\rm then }\frac{{{\rm dy}}}{{{\rm dx}}} = k.4^x {\rm where k is: ?}<br />

    Is started with trying first principles and came up with this
    <br />
\mathop {{\rm lim}}\limits_{{\rm h} \to {\rm 0}} \frac{{4^{a + h} - 4^a }}{h}<br />

    I'm stuck
    y=4^x=e^{\ln(4) x}

    So using the chain rule and the known derivative of e^xetc:

     <br />
\frac{dy}{dx}=\frac{d}{dx}\left[\ln(4) x \right]\ e^{\ln(4)x}<br />

    which you should be able to complete

    RonL
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  3. #3
    Senior Member nikhil's Avatar
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    Quote Originally Posted by Craka View Post
    How do I go about tackling the problem below?
    <br />
{\rm If }y = 4^x {\rm  then }\frac{{{\rm dy}}}{{{\rm dx}}} = k.4^x {\rm  where k is: ?}<br />

    Is started with trying first principles and came up with this
    <br />
\mathop {{\rm lim}}\limits_{{\rm h} \to {\rm 0}} \frac{{4^{a + h}  - 4^a }}{h}<br />

    I'm stuck
    if
    y=a^x then
    dy/dx=a^x(lna)
    so if
    y=4^x
    dy/dx=4^x(ln4) but its given
    dy/dx=k.4^x
    comparing both you can get the value of k=ln4
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