Find f'(x) for f(x)=sec^2x^2
I interpret that as
f(x) = ( sec(x^2) )^2
so f'(x) = 2(sec(x^2))*(sec(x^2))(tan(x^2))*(2x)
= 4x(sec(x^2) )^2 ( tan(x^2 )
where did you get the (2x) at the end from?
Chain rule:
df(x^2)/dx
=df(x^2)/d(x^2).dx^2/dx
=df(x^2)/d(x^2).2x
To find the derivative of f(x) = sec²(x²), we can use the chain rule. The chain rule states that if we have a composition of two functions, we need to differentiate the outer function and then multiply it by the derivative of the inner function.
Let's break down the function f(x) = sec²(x²) into its component functions:
g(x) = x²
h(x) = sec²(x)
First, let's calculate the derivative of g(x) using the power rule:
g'(x) = 2x
Next, let's calculate the derivative of h(x). The derivative of sec²(x) is obtained by using the chain rule:
h'(x) = 2sec(x)tan(x)
So now we have the derivatives of the inner and outer functions, g'(x) and h'(x). To find f'(x), we multiply the two derivatives:
f'(x) = g'(x) * h'(x)
= (2x) * (2sec(x)tan(x))
= 4xsec(x)tan(x)
So the derivative of f(x) = sec²(x²) is f'(x) = 4xsec(x)tan(x).