Does the Chain Rule Apply to This Derivative?

  • Thread starter Thread starter asi123
  • Start date Start date
  • Tags Tags
    Derivative
asi123
Messages
254
Reaction score
0

Homework Statement



Guys, is this right?
And if it is, from the the y' got in there?

Homework Equations





The Attempt at a Solution

 

Attachments

  • 1.jpg
    1.jpg
    6.7 KB · Views: 372
Physics news on Phys.org


asi123 said:

Homework Statement



Guys, is this right?
And if it is, from the the y' got in there?

Homework Equations





The Attempt at a Solution


You should use imageshack/other for images
 


The answer looks reasonable - perhaps multiply the two negatives.
I'm not sure what you mean by "And if it is, from the the y' got in there?"
 


Here, z and y are both functions of some other variable, perhaps x or t. If z= f(y) and y is itself a function of x, then, by the chain rule
\frac{dz}{dx}= \frac{dz}{dy}\frac{dy}{dx}

If, in particular, z= sin(1/y)= sin(y-1), and y is a function of x, then
\frac{dz}{dx}= cos(1/y)(-1/y^2)\frac{dy}{dz}
or, the ' notation,
z'= cos(1/y)(-1/y2)y'= (-cos(1/y)/y2)y'
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top