3.4

Question 1

The Chain Rule

Answer 1

If g is differentiable at x and f is differentiable at g(x), then the composite function F = f º g, defined by F(x) = f(g(x)) is differentiable at x and F’ is given by the product

F’(x) = f’(g(x)) * g’(x)

Question 2

Chain Rule in Leibniz Notation

Answer 2

If y = f(u) and u = g(x) are both differentiable functions, then

dy/dx = dy/du du/dx

Question 3

y = (x + 1)²

dy/dx =

Answer 3

y = u²

u = x + 1

dy/dx = dy/du du/dx

dy/dx = d/dx (u²) * d/dx (x+1)

dy/dx = 2u * 1 = 2u

dy/dx = 2(x+1)

dy/dx = 2x + 1

Question 4

y = e^{(x + 1)^2}

dy/dx =

Answer 4

y = e^u

u = (x+1)²

u = v²

v = x + 1

dy/dx = dy/du * du/dv * dv/dx

dy/dx = e^u * 2v * 1

dy/dx = e^{(x+1)^2} * 2(x + 1)

Question 5

The Power Rule Combined with the Chain Rule

Answer 5

if n is any real number and u = g(x) is differentiable, then

d/dx (uⁿ) = nu^n-1 du/dx

Question 6

The Power Rule Combined with the Chain Rule Written with g(x)

Answer 6

d/dx [g(x)]ⁿ = n[g(x)]^n-1 * g’(x)

Question 7

d/dx (a^x) =

Answer 7

a^x lna

Question 8

Parametric Curve

Answer 8

A curve composed of x and y values, where x and y are both functions of a third variable, t, the parameter. The parametric curve would be given by the equations

x = f(t), and y = g(t)

The curve cannot be described by the equation of form y = f(x). Thre curve may seem to break the vertical line test.

Question 9

Tangents to Parametric Curves

Answer 9

in a curve defined by the parametric equations

x = f(t) and y= g(t)

if we want to find the tangent line at a point on the curve where y is also a differentiable function of x, then the Chain Rule gives us

dy/dt = dy/dx * dx/dt

dy/dt = (dy/dt)/(dx/dt)

note how the dt’s cancel out

This allows us to find the tangent to a parametric curve without eliminating the parameter t