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Limit Definition of the Derivative We define the derivative of a function f (x) at x = a as f 0 (a) = lim h→0 f (a + h) − f (a) h f 0 (a) = lim or h→a f (h) − f (a) h−a if these limits exist! We’ll usually find the derivative as a function of x and then plug in x = a. (This allows us to quickly find the value of the derivative a multiple values of a without having to evaluate a limit for each of them.) f (x + h) − f (x) h→0 h f 0 (x) = lim f (h) − f (x) h→x h−x f 0 (x) = lim or Notation and Higher Order Derivatives The following are all different ways of writing the derivative of a function y = f (x): f 0 (x), y0 , d [f (x)] , dx df , dx dy , dx · D [f (x)] , Dx [f (x)] , f (The brackets in the third, sixth, and seventh forms may be changed to parentheses or omitted entirely.) If we take the derivative of the derivative we get the second derivative. We can also find other higher order derivatives (third, fourth, fifth, etc.). The notation for these is as follows: Order Notation Second f 00 (x), y 00 , d2 f (x), dx2 d2 f , dx2 d2 y , dx2 D2 f (x), Dx2 f (x), f Third f 000 (x), y 000 , d3 f (x), dx3 d3 f , dx3 d3 y , dx3 D3 f (x), Dx3 f (x), f Fourth nth ·· ∴ f (4) (x), y (4) , d4 f (x), dx4 d4 f , dx4 d4 y , dx4 D4 f (x), Dx4 f (x) f (n) (x), y (n) , dn f (x), dxn dn f , dxn dn y , dxn Dn f (x), Dxn f (x) When is a Function Not Differentiable? There is a theorem relating differentiability and continuity which says that if a function is differentiable at a point then it is also continuous there. The contrapositive of this theorem is that if a function is not continuous at a point then it is not differentiable there, so this gives us one way we can identify places where a function is not differentiable. While there are different types of discontinuities, all of them cause the function to not be differentiable. The other ways a function can fail to be differentiable at a point come from the limit discussed above failing to exist. This happens if the function has a sharp point or cusp, or if it has a vertical tangent line at that point. To summarize, there are three ways in which a function f (x) can fail to be differentiable at x = a: 1. f is not continuous at x = a. 2. f has a sharp point or cusp at x = a. 3. f has a vertical tangent line at x = a. 1 Derivatives of Basic Functions Power Rule: d n [x ] = nxn−1 , where n is any real number dx Derivative of a Constant: d [c] = 0 dx Trigonometric Functions: d [sin(x)] = cos(x) dx d [cos(x)] = − sin(x) dx We can use these two to find the derivatives of the remaining trig functions: d [cot(x)] = − csc2 (x) dx d [csc(x)] = − csc(x) cot(x) dx d [tan(x)] = sec2 (x) dx d [sec(x)] = sec(x) tan(x) dx Derivatives you’ll see in calculus II: Exponential Functions: Hyperbolic Trigonometric Functions: With base a: Definitions of hyperbolic sine and hyperbolic cosine: d x [a ] = ln(a) · ax dx With base e, this becomes: d x [e ] = ex dx sinh(x) = cosh(x) = ex + e−x 2 The other hyperbolic trigonometric functions defined in terms of sinh(x) and cosh(x) similar to the standard trigonometric functions. Derivatives are as follows: d [sinh(x)] = cosh(x) dx d [cosh(x)] = sinh(x) dx d [tanh(x)] = sech2 (x) dx d [coth(x)] = −csch2 (x) dx d [sech(x)] = −sech(x) tanh(x) dx d [csch(x)] = −csch(x) coth(x) dx Logarithmic Functions: With base a: d 1 1 [loga (x)] = · dx ln(a) x With base e, this becomes: d 1 [ln(x)] = dx x Inverse Trigonometric Functions: d [arctan(x)] = dx d [arcsin(x)] = dx d [arccos(x)] = dx ex − e−x 2 d −1 1 tan (x) = dx 1 + x2 d −1 1 sin (x) = √ , −1 < x < 1 dx 1 − x2 d −1 1 cos (x) = − √ , −1 < x < 1 dx 1 − x2 2 Differentiation Rules and Techniques Constant Multiple Rule: d [c · f (x)] = c · f 0 (x) dx Sum/Difference Rule: d [f (x) ± g(x)] = f 0 (x) ± g 0 (x) dx Product Rule: d [f (x) · g(x)] = f 0 (x) · g(x) + f (x) · g 0 (x) dx d Note: dx [f (x) · g(x)] 6= f 0 (x) · g 0 (x), i.e., you cannot simply take the derivatives of each function and multiply them together. There is a similar pattern for products of 3 or more functions. For instance, d [f (x) · g(x) · h(x)] = f 0 (x) · g(x) · h(x) + f (x) · g 0 (x) · h(x) + f (x) · g(x) · h0 (x) dx Quotient Rule: d f (x) f 0 (x) · g(x) − f (x) · g 0 (x) = 2 dx g(x) [g(x)] Note: d dx h f (x) g(x) i 6= f 0 (x) g 0 (x) , i.e., you cannot simply take the derivatives of each function and divide them. “Snow White and the Seven Dwarves” mnemonic: d hi = “ho dee hi minus hi dee ho over ho ho” dx ho Chain Rule: d d [(f ◦ g)(x)] = [f (g(x))] = f 0 (g(x)) · g 0 (x) = (derivative of the outside) · (derivative of the inside) dx dx Implicit Differentiation: Used on equations that define a function implicitly (as opposed to explicitly as y = f (x)). 1. Differentiate both sides of the equations w.r.t x, treating y as a function of x and using the chain rule as needed. 2. Solve for dy dx : (a) Collect all terms involving (b) then factor dy dx dy dx in the LHS (move all other terms to RHS), out of the LHS, (c) and finally divide through by LHS factor that does not involve 3 dy dx . Differentiation techniques you’ll see in calculus II: Tangents to Parametric Curves: Given a parametric curve x = f (t), y = g(t) then: dy dy dt = , dx dx dt if dx 6= 0. dt Logarithmic Differentiation: The book is using the phrase “logarithmic differentiation” to refer to two different things in this section: The first, and what most people mean when they say “logarithmic differentiation”, is a technique that can be used when differentiating a more complicated function y = f (x). Here are the steps: 1. Take the natural log of both sides: ln(y) = ln (f (x)) (we could have used any base log, but ln is a little nicer to differentiate). 2. Use properties of logs to expand the RHS as much as possible 3. Differentiate (using implicit differentiation on the LHS). 4. Solve for dy dx . 5. Substitute f (x) in for y (want final answer for dy dx in terms of x only!). The other type of problem that comes up in this section is the situation where you want to differentiate a function that looks like y = loga (something rather complicated and unpleasant) (typically the “loga ” will be “ln”). In this case, you simply use properties of logarithms to expand the RHS as much as possible, and then differentiate. dy So, there is no implicit differentiation on the LHS, solving for dx , or substituting in or multiplying by y = f (x) at the end. 4