Explanation: Let's calculate the second derivative and then find the sign of it when #x=-1#:

The answer is =2^x ln2 Let y=2^x Taking logarithms on both sides lny=ln (2^x) lny=xln2 Differentiating both sides (lny)'= (xln2)' 1/ydy/dx=ln2 dy/dx=yln2=2^xln2

x approx -0.285 Critical values are when the slope is zero. We can immediately recognize that this function has x in both the exponent and as coefficients, hence we will not be able to …

Jan 13, 2016 · According to the chain rule, the derivative is the derivative of #1/u# multiplied with the derivative of #u#.

The function will be convex (aka concave upward) if f'' (8)>0, and concave (concave downward) if f'' (8)<0 We must use the quotient rule to find this derivative.

The maximum repeats at x = pi/8+npi; n in ZZ The first one is at: 8sin (pi/8)cos (pi/8)+4-8sin^2 (pi/8) = 4sqrt2 Compute the first derivative: (d (8sin (x)cos (x)+4-8sin^2 (x)))/dx = 8 (cos^2 (x) …

Help with finding the answer? For what values of x is #f (x)= 7x^3 + 2 x^2 + 7x -2 # concave or convex? What is a solution to the differential equation #dy/dx=y#? How can you recognize …

Let #f (x) = x / (x^2+1)#, how do you use the first derivative test to determine which critical numbers, if any give relative extrema?

#f (x) = (2x^2 + x)^3 *x*x^ (-1/2) = (2x^2 + x)^3 x^ (1/2)# Use the power rule / chain rule: # (u^n)' = n (u^ (n-1))u'# Let #u = 2x^2 + x; " "n = 3; " " (u^3)' = 3 (2x^2 + x)^2 (4x + 1)# Since #f (x) = …

We have been told that the differential equation #m ( (d^2x)/ (dt^2))=-kx# is satisfied by the equation #x (t)=x_0cos (sqrt (k/mt))# Show by differentiating that this function for x (t) does in …