How do you differentiate # f(x) = sec(x^2 + 1)-tan^2x

#?

Question

How do you differentiate # f(x) =  sec(x^2 + 1)-tan^2x

#?

Cameron Alexander · Accepted Answer

df(x) =#-sec^2(x^2+1)*cos(x^2+1)*2x-2tanx*cos^2x# The decision based on the rules of differentiation of a composite function

Paisley Johnson · Answer

#2x sec(x^(2) + 1) tan(x^(2) + 1) - 2 sec^(2)(x) tan(x)#

We have: #f(x) = sec(x^(2) + 1) - tan^(2)(x)#

#=> f'(x) = (d) / (dx) (sec(x^(2) + 1)) - (d) / (dx) (tan^(2)(x))#

This function can be differentiated using the "chain rule" and the "sum rule".

Let #u = x^(2) + 1 => u' = 2x# and #v = sec(u) => v' = sec(u) tan(u)#:

#=> f'(x) = 2x cdot sec(u) tan(u) - (d) / (dx) (tan^(2)(x))#

#=> f'(x) = 2x sec(u) tan(u) - (d) / (dx) (tan^(2)(x))#

We can now replace #u# with #x^(2) + 1#:

#=> f'(x) = 2x sec(x^(2) + 1) tan(x^(2) + 1) - (d) / (dx) (tan^(2)(x))#

Now, let #u = tan(x) => u' = sec^(2)(x)# and #v = u^(2) => v' = 2 u#:

#=> f'(x) = 2x sec(x^(2) + 1) tan(x^(2) + 1) - (sec^(2)(x) cdot (2 u))#

#=> f'(x) = 2x sec(x^(2) + 1) tan(x^(2) + 1) - (2 sec^(2)(x) u)#

We can now replace #u# with #tan(x)#:

#=> f'(x) = 2x sec(x^(2) + 1) tan(x^(2) + 1) - (2 sec^(2)(x) (tan(x)))#

#=> f'(x) = 2x sec(x^(2) + 1) tan(x^(2) + 1) - 2 sec^(2)(x) tan(x)#

Adam Adkins · Answer

undefined To differentiate $ f(x) = \sec(x^2 + 1) - \tan^2(x) $, you'll need to use the chain rule and the derivative of secant and tangent functions. The derivative of $ \sec(u) $ with respect to $ x $ is $ \sec(u) \tan(u) \frac{du}{dx} $, and the derivative of $ \tan(u) $ with respect to $ x $ is $ \sec^2(u) \frac{du}{dx} $.

Now, let's differentiate each term separately:

1. Differentiate $ \sec(x^2 + 1) $:
   $$ \frac{d}{dx} \sec(x^2 + 1) = \sec(x^2 + 1) \tan(x^2 + 1) \cdot 2x $$

2. Differentiate $ \tan^2(x) $:
   $$ \frac{d}{dx} \tan^2(x) = 2 \tan(x) \sec^2(x) $$

Putting it all together:
$$ f'(x) = \sec(x^2 + 1) \tan(x^2 + 1) \cdot 2x - 2 \tan(x) \sec^2(x) $$

So, the derivative of $ f(x) = \sec(x^2 + 1) - \tan^2(x) $ is $ f'(x) = 2x \sec(x^2 + 1) \tan(x^2 + 1) - 2 \tan(x) \sec^2(x) $.