Derivatives of Trigonometric Functions

In summary, to find the second derivative of y with respect to x, first use the product rule to differentiate the original function. Then, to differentiate the result, use the quotient rule and chain rule as needed. dx/dy refers to the derivative of x with respect to y, d^2 y/dx^2 is the second derivative of y with respect to x, and dy^2/dx^2 is not a valid notation.
  • #1
Heat
273
0

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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  • #2
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).
 
  • #3
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 [itex]x(y) = 3 \sqrt{1 + y^2} / y[/itex], 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
[tex] \frac{d^2 y}{dx^2} = \frac{d\left( \frac{ dy }{ dx } \right) }{ dx } [/tex]
Of course this is possible, since dy/dx is again a function of x, for example:
[tex]y(x) = x^2, \frac{dy}{dx} = 2 x, \frac{d^2 y}{dx} = \frac{d (2x) }{dx} = 2.[/tex]

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. [itex]y = 2 x^2[/itex] s.t. [itex]y^2 = 4 (x^2)^2 = 4 x^4[/itex]. Then [itex]\frac{ d y^2 }{ d x^2 } = 8 x^2[/itex] as you can see by temporarily writing [itex]z = x^2[/itex] everywhere, deriving, and writing it back in x^2).
 
Last edited:

What are the basic trigonometric functions?

The basic trigonometric functions are sine, cosine, tangent, cotangent, secant, and cosecant.

What are the derivatives of trigonometric functions?

The derivatives of trigonometric functions are found by applying the chain rule to the basic trigonometric functions. The derivatives are: - For sine: cos(x)- For cosine: -sin(x)- For tangent: sec^2(x)- For cotangent: -csc^2(x)- For secant: sec(x)tan(x)- For cosecant: -csc(x)cot(x)

How do I solve for the derivative of a composite function involving trigonometric functions?

To solve for the derivative of a composite function involving trigonometric functions, you will need to use the chain rule. First, identify the outer function and its derivative. Then, identify the inner function and its derivative. Finally, substitute the derivatives into the chain rule formula: (outer derivative)(inner derivative).

Can you provide an example of finding the derivative of a composite function involving trigonometric functions?

Yes, for a function such as f(x) = sin(2x), the outer function is sin(x) with a derivative of cos(x), and the inner function is 2x with a derivative of 2. The derivative would be found using the chain rule as follows: f'(x) = (outer derivative)(inner derivative) = cos(2x)(2) = 2cos(2x).

What is the relationship between the derivatives of trigonometric functions and the unit circle?

The derivatives of trigonometric functions can be understood by looking at the unit circle. The derivative of sine represents the y-coordinate on the unit circle, while the derivative of cosine represents the x-coordinate. The derivatives of tangent and cotangent represent the slope of the tangent and cotangent lines on the unit circle, respectively. The derivatives of secant and cosecant represent the slope of the secant and cosecant lines on the unit circle, respectively.

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