if y=sec^(3)2x, then dy/dx=
my answer is 6 sec^3 2xtan 2x but im not sure if that is right. ?
Can someone show me if i did it correct.
good job
Sorry to bother you, can you explain the steps to me, so I can see the work and check my work. I thought i had done it wrong, I started my work and then guessed.
write it as
y = (sec (2x))^3
now use the chain rule ...
y' = 3(sec (2x)^2 (derivative of sec(2x))
= 3(sec (2x)^2(sec (2x))tan (2x))(2)
= 6(sec (2x))^3 tan (2x)
To find the derivative of y = sec^3(2x), you can use the chain rule. The chain rule states that if you have a function of the form f(g(x)), then the derivative is given by f'(g(x)) * g'(x).
First, let's differentiate the outer function f(u) = u^3 with respect to u.
f'(u) = 3u^2.
Now, let's differentiate the inner function g(x) = sec(2x) with respect to x. To do this, we can express sec(2x) in terms of sine and cosine using the identity sec(x) = 1/cos(x).
sec(2x) = 1/cos(2x).
Using the chain rule, the derivative of sec(2x) with respect to x is given by:
d/dx(sec(2x)) = d/dx(1/cos(2x)).
To find this derivative, let's use the quotient rule. The quotient rule states that if you have a function of the form f(x) = g(x) / h(x), then the derivative is given by [g'(x) * h(x) - g(x) * h'(x)] / [h(x)]^2.
In this case, g(x) = 1 and h(x) = cos(2x). Differentiating g(x) and h(x) gives:
g'(x) = 0 (since 1 is a constant)
h'(x) = -2 sin(2x).
Now applying the quotient rule:
d/dx(1/cos(2x)) = [0 * cos(2x) - 1 * (-2 sin(2x))] / [cos(2x)]^2
= 2 sin(2x) / [cos(2x)]^2.
So, the derivative of y = sec^3(2x) with respect to x is:
dy/dx = f'(g(x)) * g'(x)
= 3(sec(2x))^2 * (2 sin(2x) / [cos(2x)]^2)
= 6 sin(2x) / cos^3(2x)
= 6 sec^3(2x) tan(2x).
Therefore, your answer of 6 sec^3(2x) tan(2x) is correct.