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L'Hopital's Rule
- Thread starter jtw2e2
- Start date
You used the power rule. You are not raising a function of t to a real power, which is when the power rule applies. I found that this is a common mistake among students in Calculus.
The derivative of a^t w.r.t. to t is a^t * ln(a). You would use the power rule if you were given t^a, not a^t.
The derivative of a^t w.r.t. to t is a^t * ln(a). You would use the power rule if you were given t^a, not a^t.
BigGlenntheHeavy
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\(\displaystyle \lim_{t\to0}\frac{5^t-3^t}{t} \ gives \ one \ the \ indeterminate \ form \ \frac{0}{0}, \ hence \ the \ Marqui \ to \ the \ rescue.\)
\(\displaystyle D_t \ [5^t-3^t] \ = \ 5^tln5-3^tln3 \ and \ D_t \ [t] \ = \ 1\)
\(\displaystyle Hence \ we \ have \ \lim_{t\to0} \ [5^tln5-3^tln3] \ = \ ln5-ln3 \ = \ ln(5/3).\)
\(\displaystyle Look, \ D_t[t^2] \ = \ 2t, \ exponent \ a \ constant, \ but \ D_t[2^t] \ = \ 2^tln|t|, \ exponent \ a \ variable.\)
\(\displaystyle D_t[t^2] \ the \ hard \ way. \ Let \ y \ = \ t^2, \ then \ ln|y| \ = \ 2ln|t|, \ y \ = \ e^{2ln|t|}\)
\(\displaystyle y' \ = \ \frac{2e^{2ln|t|}}{t} \ = \ \frac{2t^2}{t} \ = \ 2t, \ Note: \ t^2 \ = \ e^{2ln|t|}\)
\(\displaystyle Now, \ D_t[2^t] \ Again \ let \ y \ = \ 2^t, \ then \ lny \ = \ tln(2) \ \implies \ y \ = \ e^{tln(2)}\)
\(\displaystyle Ergo, \ y' \ = \ ln(2)e^{tln2} \ = \ 2^tln(2), \ Again \ note: \ e^{tln(2)} \ = \ 2^t.\)
\(\displaystyle Hope \ this \ helped \ as \ it \ is \ a \ little \ confusing \ at \ first.\)
\(\displaystyle D_t \ [5^t-3^t] \ = \ 5^tln5-3^tln3 \ and \ D_t \ [t] \ = \ 1\)
\(\displaystyle Hence \ we \ have \ \lim_{t\to0} \ [5^tln5-3^tln3] \ = \ ln5-ln3 \ = \ ln(5/3).\)
\(\displaystyle Look, \ D_t[t^2] \ = \ 2t, \ exponent \ a \ constant, \ but \ D_t[2^t] \ = \ 2^tln|t|, \ exponent \ a \ variable.\)
\(\displaystyle D_t[t^2] \ the \ hard \ way. \ Let \ y \ = \ t^2, \ then \ ln|y| \ = \ 2ln|t|, \ y \ = \ e^{2ln|t|}\)
\(\displaystyle y' \ = \ \frac{2e^{2ln|t|}}{t} \ = \ \frac{2t^2}{t} \ = \ 2t, \ Note: \ t^2 \ = \ e^{2ln|t|}\)
\(\displaystyle Now, \ D_t[2^t] \ Again \ let \ y \ = \ 2^t, \ then \ lny \ = \ tln(2) \ \implies \ y \ = \ e^{tln(2)}\)
\(\displaystyle Ergo, \ y' \ = \ ln(2)e^{tln2} \ = \ 2^tln(2), \ Again \ note: \ e^{tln(2)} \ = \ 2^t.\)
\(\displaystyle Hope \ this \ helped \ as \ it \ is \ a \ little \ confusing \ at \ first.\)
daon said:You used the power rule. You are not raising a function of t to a real power, which is when the power rule applies. I found that this is a common mistake among students in Calculus.
The derivative of a^t w.r.t. to t is a^t * ln(a). You would use the power rule if you were given t^a, not a^t.
Thank you so much. That is assuming "a" is a constant, right?
Thanks Glenn, but how does one go from here:
If 5[sup:c4geqw6p]t[/sup:c4geqw6p] is a constant multiplier, doesn't it have to stay? I understand if it were a constant added to ln5 its derivative would simply be 0 but since it is multiplied, we can't just drop it, can we?
to here:BigGlenntheHeavy said:\(\displaystyle [5^tln5-3^tln3]\)
BigGlenntheHeavy said:\(\displaystyle \ = \ ln5-ln3 \\)
If 5[sup:c4geqw6p]t[/sup:c4geqw6p] is a constant multiplier, doesn't it have to stay? I understand if it were a constant added to ln5 its derivative would simply be 0 but since it is multiplied, we can't just drop it, can we?
BigGlenntheHeavy
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\(\displaystyle Anything \ raised \ to \ the \ zero \ power \ is \ 1, \ except \ 0^0.\)
\(\displaystyle Note: \ Easy \ proof, \ \frac{x^1}{x^1} \ = \ 1 \ = \ x^{1-1} \ = \ x^0, \ hence \ x^0 \ must \ = \ 1.\)
\(\displaystyle \lim_{t\to0} \ (5^t) \ = \ 1\)
\(\displaystyle Note: \ Easy \ proof, \ \frac{x^1}{x^1} \ = \ 1 \ = \ x^{1-1} \ = \ x^0, \ hence \ x^0 \ must \ = \ 1.\)
\(\displaystyle \lim_{t\to0} \ (5^t) \ = \ 1\)
BigGlenntheHeavy said:\(\displaystyle Anything \ raised \ to \ the \ zero \ power \ is \ 1, \ except \ 0^0.\)
\(\displaystyle Note: \ Easy \ proof, \ \frac{x^1}{x^1} \ = \ 1 \ = \ x^{1-1} \ = \ x^0, \ hence \ x^0 \ must \ = \ 1.\)
\(\displaystyle \lim_{t\to0} \ (5^t) \ = \ 1\)
Thank you.
It is okay to define 0^0 to be 1; many calculators do this. IMO, It makes about the same sense as defining anything to the 0th power to be 1. In group theory, g^0 is often called the "empty product" and is defined to be the multiplicative identity. I'm open to any objections, as I don't recall any reason why not to define it that way.
BigGlenntheHeavy
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\(\displaystyle daon ,\ I'll \ pass \ as \ I \ have \ been \ drawn \ into \ this \ debate \ to \ many \ times \ to \ no \ avail.\)
\(\displaystyle However, \ I'm \ curious, \ what \ does \ IMO \ mean?\)
\(\displaystyle However, \ I'm \ curious, \ what \ does \ IMO \ mean?\)