What is the Value of dy/dx When y=1 in Implicit Differentiation?

thereddevils
Messages
436
Reaction score
0

Homework Statement



If y= \ln \sqrt{xy} , find the value of dy/dx when y=1

Homework Equations





The Attempt at a Solution



\frac{dy}{dx} = \frac{1}{\sqrt{xy}} \cdot \frac{1}{2\sqrt{xy}}\cdot (x\frac{dy}{dx}+y)

\frac{dy}{dx}=\frac{1}{x} , when y=1
 
Physics news on Phys.org
Looks correct so far.
Did you try plugging in y = 1 and simplifying?
 
You answer should be a number, not a function of x. Since y= ln(xy), when y= 1, 1= ln(x). What is x?
 
HallsofIvy said:
You answer should be a number, not a function of x. Since y= ln(xy), when y= 1, 1= ln(x). What is x?

this kept me wondering again , if i differentiate first , then plug y=1 in , i get dy/dx=1/x

and now , if i plug y=1 in first , and differentiate later ,

1= ln sqrt(x)

dy/dx=1/2x

Are they supposed to end up with the same result ?
 
In all such problems, you have to differentiate first, then substitute the values.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top