What is the derivative of a^x. Try proving it.

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let us suppose y = a^x log(y) = x log(a) ; taking log on both sides y = e^(xlog(a)) ; dy/dx = log(a) * e^(xlog(a)) ; differentiating either sides dy/dx = log(a) * e^(log(a^x)) ; because alogb = log(b^a) dy/dx = log(a) * (a^x) ; because e^(log(b)) = b hence dy/dx = (a^x). log(a)
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Mother Tongue Bengali

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Math magacian

a^xloga
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Tutor

Xa^x-1
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ankit.mishra.bob@gmail.com

Take it equals to y and then take log both the sides then derivative it w.r.t. x then dy/dx=y.loga then put the value of y you have supposed and you will get the answer.
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a(pow)xloga
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Let y=a^x Taking log on both side lny=lna^x lny=xlna Differntiate on both side 1/ydy/dx=lna dy/dx=ylna Putt y=a^x dy/dx= a^xlna
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Tutor

Let y=a^x,here a is a constant, then dy/dx= a^
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Experties in mathmatics

y=a^x logy=xloga i/y dy/dx=loga dy/dx=yloga dy/dx=a^xloga
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Maths Tutor

Let F(x)= a^x F(x+h) = a^(x+h) dy/dx = lim?(h-0)??(a^(x+h)- a^x)/h? = lim?(h-0)??(a^x (a^h-1))/h? = lim?(h-0)??a^x lim?(h-0)??(a^h-1)/h? ? = a^x log?a this is known as abinitio
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