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Lesson 3-R

Lesson 3-R. Review of Differentiation Rules. Objectives. Know Differentiation Rules. Vocabulary. None new. Basic Differentiation Rules. d ---- (c) = 0 Constant dx d ---- (x ⁿ ) = nx n-1 Power Rule dx d d ---- [cf(x)] = c ---- f(x) Constant Multiple Rule

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Lesson 3-R

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  1. Lesson 3-R Review of Differentiation Rules

  2. Objectives • Know Differentiation Rules

  3. Vocabulary • None new

  4. Basic Differentiation Rules d ---- (c) = 0 Constant dx d ---- (xⁿ) = nxn-1Power Rule dx d d ---- [cf(x)] = c ---- f(x) Constant Multiple Rule dx dx d ---- (ex) = exNatural Exponent dx d 1 ---- (ln x) = -----Natural Logarithms dx x

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

  6. Derivatives of Inverse Trigonometric Functions d 1 d -1 ---- (sin-1 x) = ------------ ---- (cos-1 x) = ----------- dx √1 - x² dx √1 - x² d 1 d -1 ---- (tan-1 x) = ------------- ---- (cot-1 x) = ------------- dx 1 + x² dx 1 + x² d 1 d -1 ---- (sec-1 x) = ------------------ (csc-1 x) = ------------- dx x √ x² -1 dx 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.

  7. Other Differentiation Rules Constant to Variable Exponent Rule d ----- [ax] = ax 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.

  8. 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

  9. Product Rule Examples Find the derivatives of the following: f(x) = 7x³ ( sin x) f(t) = 5x³ex Two functions multiplied together PRODUCT RULE!! d(sin x) Hold it fixed d(7x³) Hold it fixed f’(t) = + 21x² (sin x) 7x³ (cos x) Two functions multiplied together PRODUCT RULE!! d(ex) Hold it fixed d(5x³) Hold it fixed f’(t) = + 15x² (ex) 5x³ (ex)

  10. 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.

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

  12. 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.

  13. 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!

  14. Basic Differentiation with Chain Rule d ---- (uⁿ) = nun-1 ∙ u’ General Power Rule dx d ---- (eu) = u’ ∙ euNatural Exponent dx d u’ ---- (ln u) = -----Natural Logarithms dx u

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

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

  17. Chain Rule with Product Rule Find the derivatives of the following: f(x) = ln(3x) (3x + 1)4+ 7 y = ln(4x + 1) e-6x f’(x) = 4(3x +1)³ (3) (ln(3x)) + (3x + 1)4(3/3x) u’/u n un-1 u’ 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’). y’(x) = (4/(4x + 1)) e-6x + ln(4x + 1) (-6e-6x) u’/u u’eu 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.

  18. Chain Rule with Quotient Rule Find the derivatives of the following: • d(x) = sin(ex²) /cos (9x) • g(t) = (7t + 1)4/(6t2 + 1)² u’ cos u v’ sin v cos(9x) (ex² )(2x) (cos (ex² )) – (-9 sin(9x))sin(ex² ) d’(t) = --------------------------------------------------------------------- (cos (9x))² We use chain rule in taking derivatives of both trig functions within the quotient rule nun-1 u’ nvn-1 v’ (6t² + 1)²4(7t + 1)³ (7) – 2(6t² + 1)(12t) (7t + 1)4 g’(t) = ------------------------------------------------------------------- ((6t2 + 1)²)² We used chain rule(in the general power rule) within the quotient rule.

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

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

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