D
Deleted member 4993
Guest
Find the derivative of the following function with respect to x
\(\displaystyle y = x^{{{{x}^x}^{{}^{ad \, infinitum}}\)
\(\displaystyle y = x^{{{{x}^x}^{{}^{ad \, infinitum}}\)
Hello, Subhotosh!
\(\displaystyle \text{We have: }\:y \;=\;x^y\)
\(\displaystyle \text{Take logs: }\:\ln(y) \:=\:\ln\left(x^y\right) \quad\Rightarrow\quad \ln(y) \:=\:y\cdot\ln(x)\)
\(\displaystyle \text{Differentiate implicitly: }\:\frac{1}{y}\!\cdot\!\frac{dy}{dx} \;=\;y\!\cdot\!\frac{1}{x} + \ln(x)\!\cdot\!\frac{dy}{dx}\)
\(\displaystyle \text{Multiply by }xy\!:\;\;x\!\cdot\!\frac{dy}{dx} \:=\:y^2 + xy\!\cdot\!\ln(x)\!\cdot\!\frac{dy}{dx}\)
. . \(\displaystyle x\!\cdot\!\frac{dy}{dx} - xy\!\cdot\!\ln(x)\!\cdot\frac{dy}{dx} \:=\:y^2\)
\(\displaystyle \text{Factor: }\:x\left[1-y\!\cdot\!\ln(x)\right]\frac{dy}{dx} \:=\:y^2\)
\(\displaystyle \text{Hence: }\:\frac{dy}{dx} \;=\;\frac{y^2}{x[1-y\!\cdot\!\ln(x)]}\)
\(\displaystyle \text{Therefore: }\:\frac{dy}{dx} \;=\;\frac{\left(x^{x^{x\hdots}}\right)^2} {x\left[1 - \left(x^{x^{x\hdots}}\right)\ln(x)\right]}\)
\(\displaystyle \text{We have: }\:y \;=\;x^y\)
\(\displaystyle \text{Take logs: }\:\ln(y) \:=\:\ln\left(x^y\right) \quad\Rightarrow\quad \ln(y) \:=\:y\cdot\ln(x)\)
\(\displaystyle \text{Differentiate implicitly: }\:\frac{1}{y}\!\cdot\!\frac{dy}{dx} \;=\;y\!\cdot\!\frac{1}{x} + \ln(x)\!\cdot\!\frac{dy}{dx}\)
\(\displaystyle \text{Multiply by }xy\!:\;\;x\!\cdot\!\frac{dy}{dx} \:=\:y^2 + xy\!\cdot\!\ln(x)\!\cdot\!\frac{dy}{dx}\)
. . \(\displaystyle x\!\cdot\!\frac{dy}{dx} - xy\!\cdot\!\ln(x)\!\cdot\frac{dy}{dx} \:=\:y^2\)
\(\displaystyle \text{Factor: }\:x\left[1-y\!\cdot\!\ln(x)\right]\frac{dy}{dx} \:=\:y^2\)
\(\displaystyle \text{Hence: }\:\frac{dy}{dx} \;=\;\frac{y^2}{x[1-y\!\cdot\!\ln(x)]}\)
\(\displaystyle \text{Therefore: }\:\frac{dy}{dx} \;=\;\frac{\left(x^{x^{x\hdots}}\right)^2} {x\left[1 - \left(x^{x^{x\hdots}}\right)\ln(x)\right]}\)
Denis said:Hey Subhotosh , did you read "Read before posting" ?
Please show your work next time, so we know how to help you.
There is no 'Read before posting" in this section. That is for the amateurs -- those who post in the other sections.
D
Deleted member 4993
Guest
Good one Paul .... Good one ..... :lol: :lol: :lol:
