Derivative Q/Calculus
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Derivative Q/Calculus

[From: ] [author: ] [Date: 11-09-13] [Hit: ]
f(x) = 3^x = exp[ln(3^x)] = exp[x.f(x) = ln(3).exp[x.ln(3)] = ln(3).f(3) = ln(3).3³ = 27.......
How do you solve (3^(3+h)-3^3)/(h)? The exact problem is to find the derivative of f(x)=3^x at f ' (x)=3.

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Hello,

The derivative of a power fonction is obtained by the use of the exponential expression of the power:
f(x) = 3^x = exp[ln(3^x)] = exp[x.ln(3)]

Then derivate:
f'(x) = ln(3).exp[x.ln(3)] = ln(3).3^x

Thus:
f'(3) = ln(3).3³ = 27.ln(3)

Logically,
Dragon.Jade :-)

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It depends on what you know (or are allowed to use...)!

If you know that the derivative of a^x is (a^x)*ln(a) for a > 0, then use that!

Otherwise, you'll have to use some "known results" about exponential forms.

For instance, 3^x = e^(xln3).
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keywords: Derivative,Calculus,Derivative Q/Calculus
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