Please find the Derivative of the function

y =sec^2x + tan^2x

If you don't know sec or tan derivatives, change them to 1/cos and sin/cos. I will be happy to check your work.

To find the derivative of the function y = sec^2(x) + tan^2(x), we can use the derivative rules for trigonometric functions.

First, let's rewrite the function using trigonometric identities:

y = 1/cos^2(x) + sin^2(x)/cos^2(x)

Now, let's find the derivative of each term separately:

For the first term, 1/cos^2(x), we can use the quotient rule:

d/dx (1/cos^2(x)) = (-2cos(x)(-sin(x)))/(cos^4(x)) = 2sin(x)/cos^3(x)

For the second term, sin^2(x)/cos^2(x), we can use the quotient rule as well:

d/dx (sin^2(x)/cos^2(x)) = ((2sin(x)cos(x))(cos^2(x)) - (sin^2(x))(-2cos(x)(sin(x))))/(cos^4(x)) = (2sin(x)cos^3(x) + 2sin^3(x))/(cos^4(x))

Now, we can add the derivatives of both terms to find the overall derivative:

d/dx (y) = 2sin(x)/cos^3(x) + (2sin(x)cos^3(x) + 2sin^3(x))/(cos^4(x))

Simplifying the expression further:

d/dx (y) = 2sin(x)/cos^3(x) + 2sin(x)/cos(x) + 2sin^3(x)/cos^4(x)

This is the derivative of the given function y.