CBSE (Commerce) Class 12CBSE
Share
Notifications

View all notifications
Books Shortlist
Your shortlist is empty

Solution for If X Y = E X − Y , Then D Y D X is (A) 1 + X 1 + Log X - CBSE (Commerce) Class 12 - Mathematics

Login
Create free account


      Forgot password?

Question

If \[x^y = e^{x - y} ,\text{ then } \frac{dy}{dx}\] is 

(a) \[\frac{1 + x}{1 + \log x}\]

(b) \[\frac{1 - \log x}{1 + \log x}\]

(c) not defined

(d) \[\frac{\log x}{\left( 1 + \log x \right)^2}\]

Solution

(d)

\[\frac{\log x}{\left( 1 + \log x \right)^2}\]

\[\text{ We have,} x^y = e^{x - y} \]
\[\text{ Taking log on both sides we get }, \]
\[ \Rightarrow y \log x = \left( x - y \right) \log_e e\]
\[ \Rightarrow y \log x = x - y\]
\[ \Rightarrow y \log x + y = x\]
\[ \Rightarrow y\left( 1 + \log x \right) = x\]
\[ \Rightarrow y = \frac{x}{1 + \log x}\]

\[\Rightarrow \frac{dy}{dx} = \frac{\left( 1 + \log x \right) \times 1 - x \times \left( 0 + \frac{1}{x} \right)}{\left( 1 + \log x \right)^2}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{1 + \log x - 1}{\left( 1 + \log x \right)^2}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{\log x}{\left( 1 + \log x \right)^2}\]

  Is there an error in this question or solution?

Video TutorialsVIEW ALL [1]

Solution If X Y = E X − Y , Then D Y D X is (A) 1 + X 1 + Log X Concept: Simple Problems on Applications of Derivatives.
S
View in app×