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    And the integral of a^f(x) = (a^f(x))/(f'(x) x lna) ?
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    The integral you said isn't correct. You can't apply the reverse chain rule in that way as you get a product, so when you differentiate it you use the product rule(or quotient) and you will see that it's not right.
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    hope this helps dy/dy(2^x)dx =2^xln(2) so the rule is a^x= a^xln(a)

    Proof........ dy/dx (a^x) dx....let y=a^x so ln(y)=ln(a^x) so ln(y)= xln(a) ... implicit differentiation so 1/y dy/dx =ln(a) so bringing the y up .... dy/dx = yln(a) ans as y=a^x dy/dx = a^xln(a)
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    (Original post by chhhhelsie)
    And the integral of a^f(x) = (a^f(x))/(f'(x) x lna) ?
    Your derivative formula is correct, but the integral one is not, and in fact it is one of the most common misconceptions among students - specifically, the "reverse chain rule", which many students believe is true in general, in fact only applies when f(x) is linear (of the form ax+b). Thus for example the integral of e^(x^2), letting f(x)=x^2 and a=e in other words, is not e^(x^2)/2x - in fact, this function e^(x^2) is impossible to integrate, although the proof of this result is beyond A-Level.
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    Integral (a^xln(a)) dx = a^x as its the thing you differentiated so go back
 
 
 
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