You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an alternative browser.
You should upgrade or use an alternative browser.
differentiation
- Thread starter cjbhab
- Start date
For the first one, is that the natural log e or just a variable e?. If it's the former, then no, you're incorrect.
If it's the log e, then you have a constant and the derivative is 0. If e is being used as a variable, then you are correct. If e is being used as a variable, that's a bad idea; There are many more things which could be used and this only causes confusion.
From here on out, I assume it's the log e, not a variable e.
\(\displaystyle \L\\\frac{1}{e^{-t}}=e^{t}\) You should know what the derivative of that is without much effort.
If it's the log e, then you have a constant and the derivative is 0. If e is being used as a variable, then you are correct. If e is being used as a variable, that's a bad idea; There are many more things which could be used and this only causes confusion.
From here on out, I assume it's the log e, not a variable e.
\(\displaystyle \L\\\frac{1}{e^{-t}}=e^{t}\) You should know what the derivative of that is without much effort.
Hello, cjbhab!
Has no one taught you the derivative of an exponential function?
Given: \(\displaystyle \:f(x) \:=\:e^u\)
The derivative is: \(\displaystyle \:f'(x)\:=\:e^u\cdot u'\)
In baby-talk: we get the exponential function itself, times the derivative of the expoent.
This a constant . . . The derivative is 0.
We have: \(\displaystyle \:f(x)\:=\:\frac{1}{3}e^{3x}\)
The derivative is: \(\displaystyle \:f'(x) \:=\:\frac{1}{3}\,\cdot\,e^{^{3x}}\,\cdot\,3 \:=\:e^{^{3x}}\)
Look at it again . . . We have: \(\displaystyle \;f(x)\:=\:\frac{1}{e^{-t}} \:=\:e^{^t}\)
The derivative is: \(\displaystyle \:f'(t) \:=\:e^{^t}\)
We have: \(\displaystyle \L\:f(x) \:=\:2e^{x^{\frac{1}{2}}}\)
The derivative is: \(\displaystyle \L\:f'(x) \:=\:2e^{x^{\frac{1}{2}}}\,\cdot\,\frac{1}{2}\cdot x^{-\frac{1}{2}} \:=\:\frac{e^{^{\sqrt{x}}}}{\sqrt{x}}\)
Has no one taught you the derivative of an exponential function?
Given: \(\displaystyle \:f(x) \:=\:e^u\)
The derivative is: \(\displaystyle \:f'(x)\:=\:e^u\cdot u'\)
In baby-talk: we get the exponential function itself, times the derivative of the expoent.
\(\displaystyle f(x) \:=\:3e^{^7}\)
This a constant . . . The derivative is 0.
\(\displaystyle \L f(x)\:=\:\frac{e^{^{3x}}}{3}\)
We have: \(\displaystyle \:f(x)\:=\:\frac{1}{3}e^{3x}\)
The derivative is: \(\displaystyle \:f'(x) \:=\:\frac{1}{3}\,\cdot\,e^{^{3x}}\,\cdot\,3 \:=\:e^{^{3x}}\)
\(\displaystyle f(t) \:=\:\frac{1}{e^{-t}}\)
Look at it again . . . We have: \(\displaystyle \;f(x)\:=\:\frac{1}{e^{-t}} \:=\:e^{^t}\)
The derivative is: \(\displaystyle \:f'(t) \:=\:e^{^t}\)
\(\displaystyle f(x) \:=\:2e^{^{\sqrt{x}}}\;\;\) Is this what you meant?
We have: \(\displaystyle \L\:f(x) \:=\:2e^{x^{\frac{1}{2}}}\)
The derivative is: \(\displaystyle \L\:f'(x) \:=\:2e^{x^{\frac{1}{2}}}\,\cdot\,\frac{1}{2}\cdot x^{-\frac{1}{2}} \:=\:\frac{e^{^{\sqrt{x}}}}{\sqrt{x}}\)