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 January 30th, 2010, 12:44 PM #1 Senior Member   Joined: Nov 2009 Posts: 169 Thanks: 0 Help with Differentiating F (x) = $5^{-2x^2+x}$ can some one explain what F(x) is ,is not the same as f (x), is it?
 January 30th, 2010, 12:58 PM #2 Guest   Joined: Posts: n/a Thanks: Re: Help with Differentiating It's the antiderivative (think integration). To find f(x), you need to differentiate. Since F(x) is the antiderivative, then F'(x)=f(x) Try logarithmic differentiation. $y=5^{-2x^{2}+x}$ Log of both sides: $ln(y)=ln(5^{-2x^{2}+x})$ $ln(y)=(-2x^{2}+x)ln(5)$ Derivative of both sides: $\frac{y'}{y}=ln(5)-4xln(5)$ $y'=y(ln(5)-4xln(5))$ But, remember what y is and sub it in: $y'=(5^{-2x^{2}+x})(ln(5)-4xln(5))=5^{x}(\frac{1}{25})^{x^{2}}ln(5)(1-4x)$
January 30th, 2010, 09:29 PM   #3
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Joined: Dec 2006
From: Lexington, MA

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Re: Help with Differentiating

Hello, 450081592!

They can use any letter to represent a function:

[color=beige]. . [/color]$f(x),\;F(x),\;g(x),\;P(x),\,\cdots$

Quote:
 $\text{Differentiate: }\;F(x) \:=\:5^{-2x^2+x}$

Are you aware that there is a formula for this problem?

We have an exponential function and its base is not $e.$

$\text{If }f(x) \:=\:a^u\text{, then: }\;f#39;(x) \;=\;a^u\,(\ln a)\,\left(\frac{du}{dx}\right)$

$\text{For your function: }\;F'(x) \;=\;5^{-2x^2+x}\,(\ln 5)\,(-4x+1)$

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