Derivatives of Trigonometric Functions

Heat
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Homework Statement



Find d^2y/dx^2.

y = x cos x

The Attempt at a Solution



I've been doing derivatives recently and now got into doing them with trig functions.

I thought it was,

y = x cos x = -xsinx = -xcosx

but that is the derivative of the derivative.

The problem asks for d^2/dx^2.

Can someone explain what to do in this situation?

What is the difference of dx/dy, d^2y/dx^2, dy^2/d^2x, etc...

Please and thank you. :smile:
 
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You need to use the product rule to differentiate x.cosx; i.e. d/dx(uv)=v(du/dx)+u(dv/dx). To find the second derivative you simply differentiate the result of d/dx(x.cosx).
 
Heat said:

Homework Statement



Find d^2y/dx^2.

y = x cos x
First find the first derivative. You get two terms (product rule!).

What is the difference of dx/dy, d^2y/dx^2, dy^2/d^2x, etc...

dx/dy is the derivative of x with respect to y. Writing this down sort of implies that x is a function of y, like x(y) = 3 \sqrt{1 + y^2} / y, though it's a bit against conventions to call the function x and the variable y.

d^2 y / dx^2 is the second derivative of y with respect to x, of which y is a function. It can be viewed as shorthand for
\frac{d^2 y}{dx^2} = \frac{d\left( \frac{ dy }{ dx } \right) }{ dx }
Of course this is possible, since dy/dx is again a function of x, for example:
y(x) = x^2, \frac{dy}{dx} = 2 x, \frac{d^2 y}{dx} = \frac{d (2x) }{dx} = 2.

Oh, and dy^2/dx^2 doesn't really mean anything, it's usually an error if someone writes that down. (Though it is possible to make sense of it, by viewing x^2 as a variable, e.g. y = 2 x^2 s.t. y^2 = 4 (x^2)^2 = 4 x^4. Then \frac{ d y^2 }{ d x^2 } = 8 x^2 as you can see by temporarily writing z = x^2 everywhere, deriving, and writing it back in x^2).
 
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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...
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