On what interval is the curve y=e^(-x^2) concave downward?

Question

I took the second derivative but got lost....

Also, help me with this one?
Use the Chain Rule to show that if Ø is measured in degrees then:
d/dØ(sin Ø)=(pi/180)(cos Ø)

thanks!

Answer 1

Well, you were on the right track!

It's concave downward wherever the second derivative is negative.

For the second one, let phi be the angle in radians. Then theta is a function of phi (specifically, a constant multiple), and the chain rule kicks in.

Answer 2

y' = -2x*e^(-x²)

y'' = -2*e^(-x²) + 4x²*e^(-x²) = e^(-x²) * (4x² - 2)
This is < 0 when 4x² - 2 < 0 because e^(-x²) is positive throughout. Solve the inequality and you're set.

Second question: Let f(Ø) = sin(Ø, expressed in degrees).
Then f(Ø) = sin(Ø * pi/180), using the "real" sin function (the argument is in radians).

The chain rule follows:
f'(Ø) = pi/180 * cos(Ø *pi/180) = pi/180 * g(Ø),
where g(x) = cos(x expressed in degrees).

Answer 3

y=e^(-x^2)==> y´= e^(-x^2) *(-2x) = -2* x*e^(-x^2)

y´´ = -2[e^(-x^2) + x* e^(-x^2)*(-2x)]= -2e^(-x^2)[1-2x^2]

1-2x^2 = 0 x=+- (sqrt2)/
sign of y´´
-infinity___+____-(sqrt2)/2_-_(sqrt2)/2___+___+inf


-(sqrt2)/2
If phi is expressed in degrees it is pi/180 *phi in rad (a)

so sin(phi) = sin(pi/180 phi)

You know that if u is in rad (sin u)´= cos u *(u´) chain rule

so the derivative of sin(pi/180*phi) is cos (pi/180 *phi) *pi/180

but the expression between brackets is phi mesured in degrees

So it´s proved