Question

If y is defined implicitly as a function of x by the equation x+(ln(y))^2 find dy/dx at the point (0,e)

Answer 1

here is an example from offline, not much but should help you figure it out

Answer 2

which represents a circle of radius five centered at the origin. Suppose that we wish to find the slope of the line tangent to the graph of this equation at the point (3, -4) .



How could we find the derivative of y in this instance ? One way is to first write y explicitly as a function of x . Thus,

x2 + y2 = 25 ,

y2 = 25 - x2 ,

and

$ y = \pm \sqrt{ 25 - x^2 } $ ,

where the positive square root represents the top semi-circle and the negative square root represents the bottom semi-circle. Since the point (3, -4) lies on the bottom semi-circle given by

$ y = - \sqrt{ 25 - x^2 } $ ,

the derivative of y is

$ y' = - (1/2) \big( 25 - x^2 \big)^{-1/2} (-2x) = \displaystyle{ x \over \sqrt{ 25 - x^2 } } $ ,

i.e.,

$ y' = \displaystyle{ x \over \sqrt{ 25 - x^2 } } $ .

Thus, the slope of the line tangent to the graph at the point (3, -4) is

$ m = y' = \displaystyle{ (3) \over \sqrt{ 25 - (3)^2 } } = \displaystyle{ 3 \over 4 } $ .

Unfortunately, not every equation involving x and y can be solved explicitly for y . For the sake of illustration we will find the derivative of y WITHOUT writing y explicitly as a function of x . Recall that the derivative (D) of a function of x squared, (f(x))2 , can be found using the chain rule :

$ D \{ ( f(x) )^2 \} = 2 f(x) \ D \{ f(x) \} = 2 f(x) f'(x) $ .

Since y symbolically represents a function of x, the derivative of y2 can be found in the same fashion :

$ D \{ y^2 \} = 2 y \ D \{ y \} = 2 y y' $ .

Now begin with

x2 + y2 = 25 .

Differentiate both sides of the equation, getting

D ( x2 + y2 ) = D ( 25 ) ,

D ( x2 ) + D ( y2 ) = D ( 25 ) ,

and

2x + 2 y y' = 0 ,

so that

2 y y' = - 2x ,

and

$ y' = \displaystyle{ - 2x \over 2y } = \displaystyle{ - x \over y } $ ,

i.e.,

$ y' = \displaystyle{ - x \over y } $ .

Thus, the slope of the line tangent to the graph at the point (3, -4) is

$ m = y' = \displaystyle{ - (3) \over (-4) } = \displaystyle{ 3 \over 4 } $ .

This second method illustrates the process of implicit differentiation. It is important to note that the derivative expression for explicit differentiation involves x only, while the derivative expression for implicit differentiation may involve BOTH x AND y .

The following problems range in difficulty from average to challenging

Answer 3

bye did it hep?

Answer 4

I got \[\frac{ -e + e ^{3xy} }{2 }\]

Answer 5

do you have any idea whether I did that correctly or not?

Answer 6

i don't see any equation in the question
x+(ln(y))^2
is an expression

Answer 7

sorry it should be

Answer 8

\[x + (\ln y )^{2}-e ^{xy}=0\]

Answer 9

do you know product and chain rules ?

Answer 10

yeah I can show you a picture of how I did it?

Answer 11

sure

Answer 12

ok 1 sec