My Math Forum differentiation of the form b^ax

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 August 6th, 2012, 12:27 PM #1 Senior Member   Joined: Mar 2012 From: Belgium Posts: 654 Thanks: 11 differentiation of the form b^ax hello i'm new to the concept of differentiation (dy/dx) and i read y=e^ax should become a*e^ax (for example e^3x --> 3*e^3x) but then i was stuck when i had something like y=7^3x; or something like y=b^ax so i thought lets transfer b^a to like e^A by doing this i got ln(b^a)*e^ln(b^a)x z=e^A ln(z)=A b^ax --> A*e^Ax b^ax --> ln(z)*e^ln(z)x b^ax --> ln(b^a)*e^ln(b^a)x could someone tell me if this is correct or if there would be an better way to get this ?
 August 6th, 2012, 12:49 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: differentiation of the form b^ax One way to differentiate: $y=b^{ax}$ where a and b are constants is to take the natural log of both sides: $\ln(y)=\ln$$b^{ax}$$=ax\ln(b)$ Now, implicitly differentiate with respect to x: $\frac{1}{y}\cdot\frac{dy}{dx}=a\ln(b)$ $\frac{dy}{dx}=y\cdot a\ln(b)=a\ln(b)\cdot b^{ax}$ Your result is equivalent after simplification.
 August 6th, 2012, 07:54 PM #3 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: differentiation of the form b^ax Another equivalent method but we don't have to differentiate logarithmic function this time: $y= b^{ax} = e^{ax \ln(b)}$ It is known that $\frac{d}{dx} e^x= e^x$; use chain rule: $\frac{dy}{dx}= \frac{d}{d$$ax \ln(b)$$}$$e^{ax \ln(b)}$$ \, \cdot \, \frac{d}{dx} ax \ln(b)$ $= e^{ax \ln(b)} \cdot a \ln(b) = a \ln(b) b^{ax}$ QED and we completely avoided logarithmic differentiation. Balarka .
 August 6th, 2012, 08:01 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: differentiation of the form b^ax Is exponential differentiation somehow to be preferred over logarithmic differentiation?
August 6th, 2012, 08:16 PM   #5
Math Team

Joined: Mar 2012
From: India, West Bengal

Posts: 3,871
Thanks: 86

Math Focus: Number Theory
Re: differentiation of the form b^ax

Quote:
 Originally Posted by MarkFL Is exponential differentiation somehow to be preferred over logarithmic differentiation?
Nope. Both are equivalent to me. But, I observed that many people in this forum says that they haven't taught logarithmic differentiation yet, but they do know exponential differentiation.

Balarka

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 August 6th, 2012, 08:20 PM #6 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: differentiation of the form b^ax Ah...I understand!

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