Lesson 3-R Review of Differentiation Rules

Objectives Know Differentiation Rules

Vocabulary None new

Basic Differentiation Rules d ---- (c) = 0 Constant dx d ---- (xⁿ) = nx n-1 Power Rule dx d ---- [cf(x)] = c ---- f(x)Constant Multiple Rule dx d ---- (e x ) = e x Natural Exponent dx d (ln x) = -----Natural Logarithms dx x

Trigonometric Functions Differentiation Rulesd ---- (sin x) = cos x---- (cos x) = –sin xdx d ---- (tan x) = sec² x ---- (cot x) = –csc² x dx d ---- (sec x) = sec x tan x ---- (csc x) = –csc x cot x dx Hint: The derivative of trig functions (the “co-functions”) that begin with a “c” are negative.

Derivatives of Inverse Trigonometric Functions d 1d (sin -1 x) = (cos -1 x) = dx √1 - x²dx √1 - x² d 1d (tan -1 x) = (cot -1 x) = dx 1 + x² d 1d (sec -1 x) = (csc -1 x) = dx x √ x² - 1dx x √ x² - 1 Interesting Note: If f is any one-to-one differentiable function, it can be proved that its inverse function f -1 is also differentiable, except where its tangents are vertical.

Other Differentiation Rules Constant to Variable Exponent Rule d [a x ] = a x ln a dx This is a simple example of logarithmic differentiation that we will examine in a later problem. Sum and Difference Rules d d d ---- [f(x) +/- g(x)] = ---- f(x) +/ g(x) dx dx dx In words: the derivative can be applied across an addition or subtraction. This is not true for a multiplication or a division as the next two rules demonstrate.

Product Differentiation Rule d d d ---- [f(x) g(x)] = f(x) ---- g(x) + g(x) ---- f(x) dx dx dx In words: the derivative of a product of two functions is the first function times the derivative of the second function plus the second function times the derivative of the first function. Remember product derivative’s format: Derivative (Product) = Some Product + Different Product

Product Rule Examples 1.f(x) = 7x³ ( sin x) 2.f(t) = 5x³ e x Find the derivatives of the following: Two functions multiplied together PRODUCT RULE!! Two functions multiplied together PRODUCT RULE!! Hold it fixed (sin x) Hold it fixed 7x³ d(sin x) (cos x) d(7x³) 21x² f’(t) = + Hold it fixed 5x³ d(e x ) (e x ) f’(t) = + Hold it fixed (e x ) d(5x³) 15x²

Quotient Differentiation Rule d d g(x) [f(x)] – f(x) -----[g(x)] d f(x) dx dx ---- [ ] = dx g(x) [g(x)]² In words: the derivative of a quotient of two functions is the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the square of the denominator.

– f’(t) = ( )² 1. Remember the derivative of the quotient format. Quotient Rule Example 1.f(x) = -2x² / (x³ - 1) Find the derivative of the following: Two functions in a fraction QUOTIENT RULE!! Hold it fixed (-2x²) Hold it fixed (x³ - 1) x³ Square the denominator and put it in as the denominator. d(-2x²) (-4x) 3. Put denominator in first position times derivative of numerator. (3x²) d(x³ - 1) 4. Subtract the derivative of the denominator times the numerator. 5. Simply if possible.

Differentiation Chain Rule What if I have something other than just x in one of the previous formulas? If f and g are both differentiable and F = f○g is the composite function defined by F(x) = f(g(x)) d ---- [F(x)] = f’(g(x)) g’(x) dx In words: the derivative of a composite function is equal to the derivation of the outer function (evaluated at the inner function) times the derivative of the inner function.

Differentiation Chain Rule U-Substitution version: If f & g are both differentiable and F = f○g is the composite function defined by F(x) = f(g(x)) dy dy du ---- = dx du dx The notation on above is Leibniz notation and is often referred to as u substitution. By letting u=g(x) we change f(g(x)) to f(u). Then its derivative is chained by derivative of y with respect to u multiplied by the derivative of u with respect to x. Note: we do this every time with just x involved, but we do not write it out since dx/dx = 1!

Basic Differentiation with Chain Rule d ---- (uⁿ) = nu n-1 ∙ u’General Power Rule dx d ---- (e u ) = u’ ∙ e u Natural Exponent dx d u’ ---- (ln u) = -----Natural Logarithms dx u

Trigonometric Functions Differentiation With Chain Ruled ---- (sin u) = u’∙cos u---- (cos u) = –u’∙sin udx d ---- (tan u) = u’∙sec² u ---- (cot u) = –u’∙csc² u dx d ---- (sec u) = u’∙sec u tan u ---- (csc u) = –u’∙csc u cot u dx Hint: The derivative of trig functions (the “co-functions”) that begin with a “c” are negative.

Basic Chain Rule Examples 1.f(x) = e -3x² + x 2.f(x) = csc (5x) Find the derivatives of the following: f’(x) = u’ e u where u = -3x² + x f’(x) = (-6x + 1) e -3x² + x f’(x) = u’ (-csc (u) cot (u) where u = 5x f’(x) = -5csc (5x) cot (5x) The exponent is not just x so use u-sub. Remember we never change the exponent when it is a variable We used trig rule for csc. Remember we never change what is inside the trig function even when we use the chain rule

Chain Rule with Product Rule 3.f(x) = ln(3x) (3x + 1) y = ln(4x + 1) e -6x Find the derivatives of the following: f’(x) = 4(3x +1)³ (3) (ln(3x)) + (3x + 1) 4 (3/3x) y’(x) = (4/(4x + 1)) e -6x + ln(4x + 1) (-6e -6x ) We used chain rule with u=3x+1 in the power rule term and v=3x in the ln term. This gives us 4∙(u)³∙(u’) (v) + (u) 4 ∙(v’). We used chain rule with u=4x+1 in the first term and v=-6x in the second term. Note we do not change the exponent when it has the variable in it. n u n-1 u’ u’/u u’e u u’/u

Chain Rule with Quotient Rule 1.d(x) = sin(e x² ) / cos (9x) 2.g(t) = (7t + 1) 4 / (6t 2 + 1)² Find the derivatives of the following: We use chain rule in taking derivatives of both trig functions within the quotient rule We used chain rule(in the general power rule) within the quotient rule. cos(9x) (e x² )(2x) (cos (e x² )) – (-9 sin(9x)) sin(e x² ) d’(t) = (cos (9x))² u’ cos uv’ sin v (6t² + 1)² 4(7t + 1)³ (7) – 2(6t² + 1)(12t) (7t + 1) 4 g’(t) = ((6t 2 + 1)²)² nv n-1 v’nu n-1 u’

Derivatives of Inverse Trigonometric Functions with Chain Rule d u’d -u’ ---- (sin -1 u) = (cos -1 u) = dx √1 - u²dx √1 - u² d u’d -u’ ---- (tan -1 u) = (cot -1 u) = dx 1 + u² d u’d -u’ ---- (sec -1 u) = (csc -1 u) = dx u √ u² - 1dx u √ u² - 1

Inverse Trig Examples 1.d(t) = cot -1 (e t ) 2.f(x) = sin -1 (3x²) Find the derivatives of the following: -u’ -e t d’(t) = = (u)² 1 + (e t )² We use chain rule with u=e t using inverse trig derivative. This gives us -(u’) / (1 + u²) We use chain rule with u=3x² using inverse trig derivative. This gives us (u’) / (1 - u²) ½ u’ 6x f’(x) = =  1 - (u)²  1 - (3x²)²