find second derivative of f(t)=inverse cos(4t)
To find the second derivative of the function f(t) = arccos(4t), we will need to first find the first derivative and then differentiate the result.
Step 1: Find the first derivative of f(t).
To differentiate the arccos(4t) function, we need to apply the chain rule. The derivative of arccos(u) is -1/sqrt(1 - u^2) times the derivative of u with respect to t.
So, applying the chain rule, the first derivative of f(t) = arccos(4t) is:
f'(t) = -1/sqrt(1 - (4t)^2) * d/dt(4t)
= -1/sqrt(1 - 16t^2) * 4
= -4/sqrt(1 - 16t^2).
Step 2: Find the second derivative of f(t).
To find the second derivative, we differentiate the first derivative with respect to t.
Differentiating f'(t) = -4/sqrt(1 - 16t^2) with respect to t:
f''(t) = d/dt(-4/sqrt(1 - 16t^2))
= -4 * d/dt(1 - 16t^2)^(-1/2)
Applying the chain rule, the derivative of (1 - 16t^2)^(-1/2) is:
d/dt(1 - 16t^2)^(-1/2) = (-1/2)*(1 - 16t^2)^(-3/2) * d/dt(1 - 16t^2)
= (-1/2)*(1 - 16t^2)^(-3/2) * (-32t)
= 16t/sqrt((1 - 16t^2)^3).
So, the second derivative of f(t) = arccos(4t) is:
f''(t) = -4 * 16t/sqrt((1 - 16t^2)^3)
= -64t/sqrt((1 - 16t^2)^3).
Therefore, the second derivative of f(t) = arccos(4t) is -64t/sqrt((1 - 16t^2)^3).