Partial differentiation problem

[DevonZA]

Homework Statement

Homework Equations

The Attempt at a Solution

1. If z=x+sin(##x^2##y) + ln y find ##\frac{\partial ^2z}{\partial x^2}## and ##\frac{\partial ^2z}{\partial y^2}##

2. Second order partial differentiation.

3. ##\frac {\partial z}{\partial x}## = 1 + ##cos(x^2y)## . (2x)
= ##\frac{\partial ^2z}{\partial x^2}## = ##\frac {\partial }{\partial x}## (1 + ##cos(x^2y)## . 2x)
= ##-sinx^2##y.2x.2
= ##-4xsinx^2##y

##\frac{\partial z}{\partial 2}## = ##cosx^2## + ##\frac{1}{y}##
##\frac{\partial ^2z}{\partial y^2}## = ##\frac{\partial }{\partial 2}## (##cosx^2## + ##\frac{1}{y}##)
= ##\frac{1}{y}## or ##y^{-1}##
[/B]

 

[SteamKing]

1. If z=x+sin(##x^2##y) + ln y find ##\frac{\partial ^2z}{\partial x^2}## and ##\frac{\partial ^2z}{\partial y^2}##

2. Second order partial differentiation.

3. ##\frac {\partial z}{\partial x}## = 1 + ##cos(x^2y)## . (2x)
= ##\frac{\partial ^2z}{\partial x^2}## = ##\frac {\partial }{\partial x}## (1 + ##cos(x^2y)## . 2x)
= ##-sinx^2##y.2x.2

Is this correct? what is the derivative of x² ?

 

[SteamKing]

1. If z=x+sin(##x^2##y) + ln y find ##\frac{\partial ^2z}{\partial x^2}## and ##\frac{\partial ^2z}{\partial y^2}##

2. Second order partial differentiation.

3.

##\frac{\partial z}{\partial y}## = ##cosx^2## + ##\frac{1}{y}##

Is ##\frac{\partial cos(x^2y)}{\partial y}=cos(x^2) ## ?

that would mean that ##\frac{d cos(θ)}{dθ} = -sin(1)##

You should review the rules of differentiation and know them thoroughly before doing partials.

 

[DevonZA]

= ##\frac{\partial ^2z}{\partial x^2}## = ##\frac {\partial }{\partial x}## (1 + ##cos(x^2y)## . 2x)
= -2sinxy.2x.2
= -8xsinxy
Is that correct?

Partial differentiation is new to me..clearly

 

[D H]

You presumably wouldn't have a problem were you asked to compute the ordinary derivatives ##\frac{d^2}{dx^2}\left(x+\sin(x^2 b)+\ln b\right)## and ##\frac{d^2}{dy^2}\left(a + \sin(a^2 y) + \ln y\right)##, where ##a## and ##b## are some constants. Now replace ##a## and ##b## with ##x## and ##y## and voila! you have your partial derivatives. When computing the partial derivative of some function with respect to one of the independent variables, you simply treat the other independent variables as if they were constants.

 

[SteamKing]

= ##\frac{\partial ^2z}{\partial x^2}## = ##\frac {\partial }{\partial x}## (1 + ##cos(x^2y)## . 2x)
= -2sinxy.2x.2
= -8xsinxy
Is that correct?

Partial differentiation is new to me..clearly

Yes, but I'm afraid the problem goes much deeper, back to taking ordinary derivatives.

What would be the derivative of this expression (no partials): ##\frac{d}{dx}cos(x^2y)## ?

 

[DevonZA]

Yes, but I'm afraid the problem goes much deeper, back to taking ordinary derivatives.

What would be the derivative of this expression (no partials): ##\frac{d}{dx}cos(x^2y)## ?

##\frac{d}{dx}cos(x^2y)## = -sin2x

 

[RUber]

##\frac{d}{dx}cos(x^2y)## = -sin2x

You need to use the chain rule.
##\frac{d}{dx} f(g(x)) = f'(g(x) ) g'(x) ##
So
##\frac{d}{dx} cos(x^2) = -sin(x^2) \frac{d}{dx} x^2 ##
Similarly,
##\frac{d}{dx} cos(x^2y) = -sin(x^2y) \frac{d}{dx} x^2y ##

 

[DevonZA]

Thanks again RUber that makes sense. I'm out at dinner currently but I'll attempt to answer the question again when I'm home later.

 

[DevonZA]

Okay I have no idea if I am doing this right but let me know..

 

[SteamKing]

Okay I have no idea if I am doing this right but let me know..

You can't have "sin" or "cos" standing by themselves. These are functions which must show an argument of some kind.

You've calculated ##\frac{\partial z}{\partial x} ## correctly, but from that point on, it seems you are having trouble applying the chain rule correctly again.

Again, I repeat, you should not be working partial derivative problems unless and until you understand derivatives of a single variable thoroughly.

 

[RUber]

##\frac{d}{dx} x^2 y ## implies that you treat y as a constant. You can't just drop it out. It isn't like taking the derivative of x^2 +y.
Take your time. Be careful with your chain rule. You can get it done.

 

[DevonZA]

You can't have "sin" or "cos" standing by themselves. These are functions which must show an argument of some kind.

You've calculated ##\frac{\partial z}{\partial x} ## correctly, but from that point on, it seems you are having trouble applying the chain rule correctly again.

Again, I repeat, you should not be working partial derivative problems unless and until you understand derivatives of a single variable thoroughly.

I had a feeling that the trig functions were wrong..
I struggle with derivatives so yes I need more practice but time is not on my side. I'll try again

 

[DevonZA]

##\frac{d}{dx} x^2 y ## implies that you treat y as a constant. You can't just drop it out. It isn't like taking the derivative of x^2 +y.
Take your time. Be careful with your chain rule. You can get it done.

So I haven't calculated ## \frac {\partial{z}}{\partial{x}}## correctly then?

##\frac{d}{dx} x^2y = 2xy## ?

 

[RUber]

So I haven't calculated ## \frac {\partial{z}}{\partial{x}}## correctly then?

##\frac{d}{dx} x^2y = 2xy## ?

That's right.

 

[DevonZA]

Please see the attachment. If you could help me with the next step from here please, the trig functions are confusing me.

 

[RUber]

That looks like you might need to use the product rule. Then for the part of the product that has d/dx cos(x^2y), use the chain rule again.
You have already shown that you can use these methods. Take your time and do them one by one.

 

[DevonZA]

Final answer attached. Thanks to all who helped.