How do I use implicit differentiation to solve these two questions?

[Ry122]

Im having trouble solving these two questions. I don't know where to start so I can't give an attempt at either of them. Please tell me how to do the full question if you can cause i can't check back until morning then i have to go to an exam.

http://www.users.on.net/~rohanlal/one.jpg
http://www.users.on.net/~rohanlal/two.jpg

 

[dirk_mec1]

First differentiate the LHS remembering that a dx will pop up then the same for the RHS then divide to find dy/dx.

 

[Nick89]

Simply differentiate both sides like you are used to, but keep in mind that y is not a constant! The derivative of y is \frac{dy}{dx} so you should leave that in the equation. Then you will find an equation which you can solve for dy/dx.

I don't know what dirk_mec means by 'a dx will pop up' though. Perhaps take a look at the following example (to not give away the answer I use a different example):

y \sin x = x^3 + \cos y
\frac{d}{dx}(y \sin x) = \frac{d}{dx}(x^3) + \frac{d}{dx}(\cos y)
(\sin x)\frac{dy}{dx} + y \cos x = 3x^2 - (\sin y) \frac{dy}{dx}
(\sin x + \sin y) \frac{dy}{dx} = 3x^2 - y \cos x
\frac{dy}{dx} = \frac{ 3x^2 - y \cos x}{\sin x + \sin y}

Note the use of the product rule on the LHS in the third line!This is called implicit differentiation. Often you can rewrite a function to read 'y = f(x)' (where f is some function). But in this case this is impossible, so you have to use implicit differentiation.

 

[Ry122]

Here are my attempts:
x^2=ln(x+y)
2x=(1/x)(1/y)dy/dx
2(x^2)y=dy/dx

e^(xy)=x+4
u=xy
u'=(1)(1)(dy/dx)
(1)(1)(dy/dx)e=1
(dy/dx)=1/e

They're both wrong.

 

[Mute]

Here are my attempts:
x^2=ln(x+y)
2x=(1/x)(1/y)dy/dx
2(x^2)y=dy/dx

\ln(x+y) = \frac{1}{x +y}\left(1 + \frac{dy}{dx}\right)

In general, the derivative of \ln f(x) is \frac{1}{f(x)}\frac{df}{dx}, as per the chain rule. In this case, f(x) = x + y(x) (y is a function of x).

e^(xy)=x+4
u=xy
u'=(1)(1)(dy/dx)
(1)(1)(dy/dx)e=1
(dy/dx)=1/e

They're both wrong.

This is just an application of the chain rule again, keeping in mind that y is a function of x. There's no need to introduce u.

\frac{d}{dx}e^{xy} = e^{xy}\frac{d}{dx}(xy) = e^{xy}\left(y + x\frac{dy}{dx}\right)

 

[Ry122]

So for ln(x+y)
its 1/(x+y) multiplied by the derivative of what's inside the parenthesis?

 

[Nick89]

So for ln(x+y)
its 1/(x+y) multiplied by the derivative of what's inside the parenthesis?

Yes.

And for e^{xy} it's \frac{d}{dx} e^{xy} = e^{xy} \frac{d}{dx}(xy)
(don't forget the product rule!)

So in words, the derivative of exp(f(x)) is simply exp(f(x)) multiplied by the derivative of f(x).So to get you started with the second one, this is the first step:
e^{xy} = x + 4
e^{xy} \frac{d}{dx}(xy) = \frac{d}{dx}(x+4)
e^{xy} \left( y + x \frac{dy}{dx} \right) = 1
...