Explicit function Example Which of the following are not explicit functions.

y = stuff with x y = x2 + sin(ex cos( x2 + 1)) Given the input we have a clear method to get p the output. x2 + y = ex + 17

sin(x + ey)=x2 - cos(y)-y Implicit relationship y = 5t2 + cos t - e-t stuff with x and y = other stuff with x and y x2 + y2 = 25

Given the “input” there is some relationship to x3 + xy - 2y2 = 7 the “output” but it might not be possible to express this relationship explicitly. x = 5y2 - cos y Observation: If f(x)=g(x), then f0(x)=g0(x). Big idea: An implicit relationship is still a relationship, in other words there is some Example Let f(✓)=cos2 ✓ + sin2 ✓ and g(✓)=1. function, we just don’t know it. So treat y as a Verify that f0(✓)=g0(✓). function of x and take the derivative of both sides with respect to x and use the chain rule dy (hint: the derivative of y with respect to x is dx).

2 x2 + y2 = 25 x2 + y(x) = 25 Taking the derivative of both sides with respect to x would give !

1 dy 2x + 2 y(x) = 0. dx Now rearrange to get dy 2x x =- =- . dx 2y(x) y

This process of finding the rate of change of two variables connected through an implicit relationship is known as implicit differentiation. Example Find the slope of the tangent line to Highlights of implicit differentiation the circle of radius 5 at the point (3, 4) by using (i) implicit differentiation; and (ii) writing y 1. Take derivative of both with respect to one explicitly as a function of x and using explicit of the variables. differentiation. How do your answers compare? 2. Make sure to apply chain rule correctly. 3. After taking derivatives carefully rearrange and solve for desired derivative.

dy 3 xy Example Find dx given y + e = 1. Notational note: A normal lie to a curve is a line which passes dy through the point and is perpendicular to the = derivative at a particular point (a, b). tangent line. dx (a,b) Example Find the tangent line to the implicitly Example Find the normal line to the implicitly 3 3 defined curve x3 + y3 = 9xy at (2, 4). (Find the defined curve x + y = 9xy at (2, 4). dy slope in two ways: (i) explicitly solve for dx and then evaluate; (ii) after taking derivative dy evaluate then solve for dx (2,4)).) d2y Example Given x3 + y3 = 9xy find . dx2 (2,4) Inverse functions If y = mx + b is tangent to y = f(x) at (a, f(a)), then we have that x = my + b is tangent to y = f-1(x) at (f(a), a).

1 y = f(x) x = my + b becomes y = (x - b). m So we can now relate the slope of the tangent line at (a, f(a)) for y = f(x) to the slope of the tangent line at (f(a), a) for y = f-1(x) by noting they are reciprocals. y = f-1(x)

Rule for derivative of inverse

d 1 Functions are related to their inverses by f-1(x) = -1 flipping across the line y = x. dx f0 f (x) Observation: Tangent lines will also flip across the line y = x. Implicit function approach Example Given f(x)=x3 + 4x + 5 and g(x)=f-1(x), find the tangent line to y = g(x) at x = 10. y = f-1(x)

f(y)=x

d d f(y) = (x) dx dx dy f0(y) = 1 dx dy 1 = dx f0(y) d 1 f-1(x) = -1 dx f0 f (x) Exponential functions and logarithms Applying the chain rule we have

d f0(x) f(x)=ex f-1(x)=ln x ln f(x) = dx f(x) eaeb = ea+b ln(ab)=ln(a)+ln(b) ! Example Find the following derivatives. (ea)b = eab ln(ab)=b ln(a) ! d ln(sec x) = ea dx = ea-b ln a = ln(a)-ln(b) eb ! b x ln(x) d ln(e )=x e = x (if x>0) ln(x100) = ! dx 1 e = e ln(e)=1 ! d e0 = 1 ln(1)=0 100 ln(x) = ! dx ! d 3 If y = ln x, then x = ey and implicit ln(x - 3x + 2) = y dx differentiation gives 1 = e y0 = xy0 which rearranging gives d ln(sec x + tan x) = dx d 1 ln(x) = dx x We know how to take the derivative of ex, more Example Find the derivative of f(x)=xx by generally we can take the derivative of ax completing the following:

d d d d d x ax = (eln(a))x = e(ln a)x xx = eln x = dx dx dx dx dx =(ln a)e(ln a)x =(ln a)ax.

d What is xx ? dx If we think of x as being in the base and • d k k-1 use the rule dx(x )=kx we would have x xx-1. · If we thing of x as being in the exponent • d x x and use the rule dx(a )=(ln a)a we would have (ln x)xx.

Which of the above two ideas do we use? d Logarithmic differentiation Example Find ex(x2 + 1)cos x . dx Properties of logarithms are useful to pull apart things which involve multiplication, division, and exponentiation. So when faced with the task of taking a derivative of a function involving these ideas we can: 1. take the logarithm; 2. use properties of logarithms to simplify/separate the expression; 3. take the derivative of each (small) piece. To make this work we note: d f0(x) d f(x) = f0(x)=f(x) = f(x) ln f(x) dx · f(x) dx d d f(x) = f(x) ln f(x) dx dx 2 d (x3 + x)(sin x)x Example Find . dx p 2 ✓ sec x x + 4 ◆ Inverse trigonometric functions

y = sin(x) y = tan(x) y = sec(x)

y = arcsin(x)=sin-1(x) y = arctan(x)=tan-1(x) y = arcsec(x)=sec-1 x Example Find the following: Goal: Find the derivatives of the inverse d ln(1 + x2) = trigonometric functions. dx y = arctan x d x = tan y arctan(sin x) = dx dy 1 = sec2 y dx d arctan(x2) = tan2 y + 1 = sec2 y dx dy dy 1 =(tan2 y + 1) =(x2 + 1) d dx dx arctan(px) = dx d 1 arctan x = dx 1 + x2 d arctan(ex) = dx Example Find the tangent lines to y = arctan x Example For time t > 0 the position of a at x = 0 and at x = 1. particle is given by s(t)=arctan(t2). Determine when the acceleration of this particle is zero. y = arcsin x y = arcsec x

x = sin y x = sec y

dy dy 1 = cos y 1 = sec y tan y dx dx

2 2 2 cos2 y + sin2 y = 1 so cos y = 1 - sin2 y tan y + 1 = sec y so tan y = sec y - 1 q p 2 dy 2 dy 2 dy dy 1 y y 1 x x 1 1 = 1 - sin y = 1 - x2 = sec sec - = - dx dx dx dx q p p p d 1 d 1 arcsec x = arcsin x = dx p 2 dx p1 - x2 x x - 1 (Note the above holds only when x>1.) d Example Find tan(arcsec x) . dx