If f(u)=u+1/(cos^2u) , u=g(x)=pi(x), x=1/4 , find the derivative of f(g(x)). I can't seem to get this one, step by step please?
let u = π/x
v = u + 1/(cos^2 u)
OR is it v = (u+1)/(cos^2 u)
I will assume it is the way you typed it, then
v = u + (cos u)^-2
dv/du = 1 - 2(cosu)^-3 (-sinu)
= 1 + 2sinu/cos^3 u
u = πx
when x = 1/4, u = π/4
dv/du = 1 + 2sin(π/4)/cos^3 (π/4)
= 1 + 2tan(π/4) /cos^2 (π/4)
= 1 + 2(1)/(1/2) = 5
du/dx = π
then dv/dx = dv/du * du/dx
= 5 * π
= 5π
better check my work
This looks right! Thanks so much!!
But I was wondering- is it possible to use the (f•g)'= f'(g(X))*g(X)' and still obtain the same answer? For the composite rule, we sub u into the dv/du equation and then take the derivative of that ONCE. Then we multiply that entire function by pi which is the derivative of pix and then sub in the X=1/4. I don't think I did it right and get 5pi. Could you please check the answer again in this way as well please?
To find the derivative of f(g(x)), we need to apply the chain rule. The chain rule states that if you have a composite function, f(g(x)), where f(u) and g(x) are both differentiable, then the derivative of f(g(x)) with respect to x is given by:
(d/dx) f(g(x)) = f'(g(x)) * g'(x)
Let's break it down step by step.
Step 1: Find g(x) and g'(x)
We are given that g(x) = π(x), where π(x) represents the floor function. Since x = 1/4, we have g(x) = π(1/4). The floor function π(x) rounds down the number x to the nearest integer, so π(1/4) = 0.
To find g'(x), we differentiate g(x) with respect to x. Since g(x) is a constant (0 in this case), the derivative g'(x) is zero.
Therefore, g(x) = 0 and g'(x) = 0.
Step 2: Find f(u) and f'(u)
We are given that f(u) = u + 1/cos^2(u). To find f'(u), we differentiate f(u) with respect to u.
Using the rules of differentiation, we have:
f'(u) = 1 + (-2/cos^3(u)) * (-sin(u))
= 1 + 2sin(u)/cos^3(u)
Step 3: Apply the chain rule
Now that we have g(x), g'(x), f(u), and f'(u), we can apply the chain rule to find (d/dx) f(g(x)).
(d/dx) f(g(x)) = f'(g(x)) * g'(x)
Substituting the values we found, we have:
(d/dx) f(g(x)) = f'(g(x)) * g'(x)
= [1 + 2sin(g(x))/cos^3(g(x))] * g'(x)
= [1 + 2sin(0)/cos^3(0)] * 0
= [1 + 0/1] * 0
= 1 * 0
= 0
Therefore, the derivative of f(g(x)) is 0.