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Confused with when to use chain rule

  1. Oct 26, 2011 #1
    Just some general questions as I'm confused with when to use chain rule when not to.

    For instance, to find the derivative of e^sqrt(x), the right answer is to use chain rule to get e^sqrtx*the derivative of sqrt(x). BUT, isn't there a formula that: d/dx K^x = In(K)*K^x? K for constant and x for differentiable function. So why I can't use it to get e^sqrt(x)=Ine*e^sqrt(x)? AND, isn't d/dx(e^x) = e^x?? I'm completely confused.. Midterm tomorrow.. really need your help.. Thanks a lot!
     
  2. jcsd
  3. Oct 26, 2011 #2

    Mark44

    Staff: Mentor

    No, there isn't. There is no function named In. Are you thinking ln (LN) for the natural logarithm?
    d/dx(ex) = ex, but what about d/dx(ef(x))? For that you need the chain rule.

    Let u = f(x).
    d/dx(ef(x)) = d/dx(eu) = d/du(eu) * du/dx = eu * du/dx.
     
  4. Oct 26, 2011 #3
    I don't know what you mean by In(K). Do you mean ln(K) (natural logarithm of K)? Assuming that's ln:

    Yes, but x there isn't a function, it's the variable you're deriving in relation to! The formula for the derivate function of an exponential is:

    [tex]\frac{d}{dx}K^{f(x)} = \frac{df(x)}{dx}ln(K)K^{f(x)}[/tex]

    The special case when f(x) = x gives that equation you wrote, because [tex]\frac{df(x)}{dx} = 1[/tex]. But when f(x) isn't x, [tex]\frac{df(x)}{dx}[/tex] won't be 1.

    To deduce this equation you need to do the chain rule.
     
  5. Oct 26, 2011 #4

    LCKurtz

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    The short answer is you have to use the chain rule whenever the argument isn't just a simple x. So you don't need it for ex. But you need it for e to anything more complicated than that: [itex]e^{2x},\ e^{\sqrt x},\ e^{-x}[/itex] etc.

    It's the same with any function. You don't need it for sin(x) but you do for the sine of anything else:[itex]\sin(2x),\ sin(-x),\ sin(x^3)[/itex]etc.

    You could be safe and always use it. For example, if you use it on sin(x) you will just get the derivative as [itex]\cos(x)\cdot 1[/itex]. The extra 1 is correct and doesn't hurt anything.
     
  6. Oct 26, 2011 #5
    Thank you so much for all of your helps! Thanks!!
     
  7. Oct 26, 2011 #6
    I thought you technically always use the chain rule anyway.

    For instance,
    [tex]\frac{d}{dx}y=x^2[/tex]
    [tex]\frac{dy}{dx}=2x(x')[/tex]
    [tex]\frac{dy}{dx}=2x[/tex]

    You are still using the chain rule, it's just that x' is 1 because the function is being differentiated with respect to x.
     
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