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Math Help - Intro Calc

  1. #1
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    Intro Calc

    Find the derivative using first principles

    y= 3
    sqrtx

    Can someone help me solve this tks
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by johntuan View Post
    Find the derivative using first principles

    y= 3
    sqrtx

    Can someone help me solve this tks
    Differentiating using first principles means they want you to use the limit definition of a derivative. There are two equivalent definitions that are used, which one did you learn?

    Definition 1:

    For a function f(x), it's derivative, f'(x), at an arbitrary point x is given by:

    f'(x) = \lim_{h \to 0} \frac {f(x + h) - f(x)}{h}



    Defintion 2:

    For a function f(x), it's derivative, f'(x), at an arbitrary point a is given by:

    f'(a) = \lim_{x \to a} \frac {f(x) - f(a)}{x - a}


    I think the first definition is what you want, so just plug the pieces into the formula and find the limit. (Do you need help with that?)
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  3. #3
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    i learnt both ways but can you solve it using definition 1 please. And yes i need help, when i plug it in I keep getting stuck.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by johntuan View Post
    i learnt both ways but can you solve it using definition 1 please
    f(x) = \frac {3}{ \sqrt {x}}

    \Rightarrow f'(x) = \lim_{h \to 0} \frac {f(x + h) - f(x)}{h}

    \Rightarrow f'(x) = \lim_{h \to 0} \frac { \frac {3}{ \sqrt {x + h}} - \frac {3}{ \sqrt {x}}}{h} ...........................................Combine the fractions in the numerator

    \Rightarrow f'(x) = \lim_{h \to 0} \frac {\frac {3 \sqrt {x} - 3 \sqrt {x + h}}{\sqrt {x} \sqrt {x + h}}}{h} ..............................................Now combined the fractions once again

    \Rightarrow f'(x) = \lim_{h \to 0} \frac {3 \sqrt {x} - 3 \sqrt {x + h}}{h \sqrt {x} \sqrt {x + h}} .....................................Now multiply by the conjugate of the numerator over itself (This is a standard approach).

    \Rightarrow f'(x) = \lim_{h \to 0} \frac {3 \sqrt {x} - 3 \sqrt {x + h}}{h \sqrt {x} \sqrt {x + h}} \cdot \frac {3 \sqrt {x} + 3 \sqrt {x + h}}{3 \sqrt {x} + 3 \sqrt {x + h}} ...........Now simplify this

    \Rightarrow f'(x) = \lim_{h \to 0} \frac {9x - 9(x + h)}{h \left(  \sqrt {x} \sqrt {x + h} \right) \left( 3 \sqrt {x} + 3 \sqrt {x + h} \right)} ...........Simplify more

    \Rightarrow f'(x) = \lim_{h \to 0} \frac {-9h}{h \left( \sqrt {x} \sqrt {x + h} \right) \left( 3 \sqrt {x} + 3 \sqrt {x + h} \right)} ............Now we see that the h's cancel

    \Rightarrow f'(x) = \lim_{h \to 0} \frac {-9}{ \left( \sqrt {x} \sqrt {x + h} \right) \left( 3 \sqrt {x} + 3 \sqrt {x + h} \right)} ...............Now there's no problem when we take the limit

    \Rightarrow f'(x) = \frac {-9}{6 \sqrt {x} x}

    \Rightarrow f'(x) = - \frac {3}{2} x^{- \frac {3}{2}}
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  5. #5
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    ok thanks i see where i messed up
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