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How do you differentiate given #y = (sec)^2 x + (tan)^2 x#?

Answer 1

Isaiah Allen

#y' = 4sec^2xtanx#

#y = sec^2x + tan^2x# #y = secx secx + tanx tanx#

#y' = (dy)/(dx) = (secxtanx)secx+secx(secxtanx)+ (sec^2x)tanx+tanx(sec^2x)#

#y' = sec^2xtanx + sec^2xtanx+sec^2xtanx+sec^2xtanx#

#y' = 4sec^2xtanx#

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Answer 2

Isaiah Allen

To differentiate y = sec^2(x) + tan^2(x), you can use the chain rule and the derivatives of sec^2(x) and tan^2(x). The derivative of sec^2(x) is 2sec(x)tan(x)sec(x), and the derivative of tan^2(x) is 2tan(x)sec^2(x). Therefore, the derivative of y with respect to x is:

dy/dx = d/dx(sec^2(x)) + d/dx(tan^2(x)) dy/dx = 2sec(x)tan(x)sec(x) + 2tan(x)sec^2(x) dy/dx = 2sec^2(x)tan(x) + 2tan(x)sec^2(x) dy/dx = 2sec^2(x)tan(x) + 2tan(x)sec^2(x)

So, the derivative of y = sec^2(x) + tan^2(x) is dy/dx = 2sec^2(x)tan(x) + 2tan(x)sec^2(x).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.