Find the second derivative of the function: y=e^-2t*cos^2(t+(pi/4))
To find the second derivative of the given function, first, let's find the derivative of y with respect to t. Then we'll take the derivative of that result again.
Given function: y = e^(-2t) * cos^2(t+(π/4))
Step 1: Find the derivative of y with respect to t.
To find the derivative, we can use the product rule and the chain rule.
Using the product rule, the derivative of the product of two functions u and v can be found using the formula: (u*v)' = u'v + uv'
Let's define u and v as:
u = e^(-2t)
v = cos^2(t + (π/4))
u' (derivative of u with respect to t) can be found using the chain rule:
u' = (e^(-2t))'
The derivative of e^(-2t) with respect to t can be computed by multiplying by the derivative of -2t with respect to t. The derivative of -2t is -2.
So, u' = (-2)e^(-2t) = -2e^(-2t)
To find v', we apply the chain rule:
v' = (cos^2(t + (π/4)))'
To find the derivative of cos^2(t + (π/4)), we consider the composite function f(g(t)), where f(x) = x^2 and g(t) = cos(t + (π/4)). By applying the chain rule, we have:
(f(g(t)))' = f'(g(t)) * g'(t)
Using f(x) = x^2, we have f'(x) = 2x
Therefore, f'(g(t)) = 2 * cos(t + (π/4))
The derivative of g(t) = cos(t + (π/4)) can be computed using the chain rule, where the derivative of cos(x) is -sin(x):
g'(t) = (cos(t + (π/4)))' = -sin(t + (π/4))
Thus, v' = f'(g(t)) * g'(t) = 2 * cos(t + (π/4)) * (-sin(t + (π/4)))
Now we can find the derivative of y with respect to t:
y' = u'v + uv'
= (-2e^(-2t)) * cos^2(t + (π/4)) + (e^(-2t)) * 2 * cos(t + (π/4)) * (-sin(t + (π/4)))
Simplifying:
y' = -2e^(-2t) * cos^2(t + (π/4)) - 2e^(-2t) * cos(t + (π/4)) * sin(t + (π/4))
Step 2: Find the second derivative of y with respect to t.
To find the second derivative, we just need to differentiate the expression obtained in step 1, with respect to t.
Differentiating y' with respect to t will give us:
y'' = (-2e^(-2t) * cos^2(t + (π/4)) - 2e^(-2t) * cos(t + (π/4)) * sin(t + (π/4)))'
Now, we need to differentiate each term using the product rule, chain rule, and derivative rules.
Differentiating the first term:
(-2e^(-2t) * cos^2(t + (π/4)))'
Using the product rule, the derivative of the product of two functions a and b can be found using the formula: (ab)' = a'b + ab'
Let's define a and b as:
a = -2e^(-2t)
b = cos^2(t + (π/4))
a' (derivative of a with respect to t) can be found using the chain rule:
a' = (-2e^(-2t))'
The derivative of -2e^(-2t) with respect to t is found by multiplying by the derivative of -2t with respect to t, which is -2.
So, a' = (-2)(-2e^(-2t)) = 4e^(-2t)
To find b', we apply the chain rule:
b' = (cos^2(t + (π/4)))'
Using the chain rule, we find:
(b^2)' = 2cos(t + (π/4))(cos(t + (π/4)))'
To find the derivative of cos(t + (π/4)), we use the chain rule:
(cos(t + (π/4)))' = -sin(t + (π/4))
Applying the above result to (b^2)', we get:
(b^2)' = 2cos(t + (π/4)) * (-sin(t + (π/4))) = -2cos(t + (π/4)) * sin(t + (π/4))
Now let's find the second derivative y'' by substituting the derivatives we've computed:
y'' = (-2e^(-2t) * cos^2(t + (π/4)))' - 2e^(-2t) * cos(t + (π/4)) * sin(t + (π/4))
= (4e^(-2t) * cos^2(t + (π/4))) + (-2cos(t + (π/4)) * sin(t + (π/4))) - 2e^(-2t) * cos(t + (π/4)) * sin(t + (π/4))
Simplifying further, we get:
y'' = 4e^(-2t) * cos^2(t + (π/4)) - 2cos(t + (π/4)) * sin(t + (π/4)) - 2e^(-2t) * cos(t + (π/4)) * sin(t + (π/4))
Therefore, the second derivative of the given function is:
y'' = 4e^(-2t) * cos^2(t + (π/4)) - 2cos(t + (π/4)) * sin(t + (π/4)) - 2e^(-2t) * cos(t + (π/4)) * sin(t + (π/4))