what is the derivative of tan x square using first principle
Answers
Answer:
suppose it’s allowed as “first principles” to use the definition of derivative:
ddxtan(x2)=limh→0tan((x+h)2)−tan(x2)h.
Hopefully it’s also allowed as “first principles” to use the trigonometric identity that
tanα=sinαcosα.
Notice that, for any a,b ,
tana−tanb=sinacosa−sinbcosb
=sinacosb−sinbcosacosacosb.
If we again allow as “first principles” the trigonometric identity that
sin(a−b)=sinacosb−sinbcosa ,
this means that
tana−tanb=sin(a−b)cosacosb,
and in particular,
tan((x+h)2)−tan(x2)=sin((x+h)2−x2)cos((x+h)2)cos(x2).
Now write (x+h)2−x2 as (x+h+x)(x+h−x)=(2x+h)h .
Putting this all together,
ddxtan(x2)=limh→0sin((2x+h)h)hcos((x+h)2)cos(x2).
While we have simplified the original expression quite a bit, it seems we hit a bit of a roadblock right now. As it turns out, we can’t get rid of the h in the denominator unless we use one of those identities of the type
limt→0sintt=1,
which i’m not sure is allowed as “first principles”. I really don’t see any other way around this short of deriving that identity from whatever it is that is considered “first principles”.
So, assuming that this identity is part of what we consider “first principles”, how do we proceed? Well, the usual trick is to observe that (2x+h)h→0 when h→0 , and thus
limh→0sin((2x+h)h)(2x+h)h=1.
So if we get a 2x+h into the denominator, we can finally solve the limit. Thus we multiply and divide by 2x+h :
ddxtan(x2)=limh→0sin((2x+h)h)(2x+h)h2x+hcos((x+h)2)cos(x2).
Now this is a product of two expressions whose limits are known. The first quotient goes to 1 as we noticed above. The other goes (using the fact that the functions involved are continuous, which hopefully is also included among these “first principles”) to
limh→02x+hcos((x+h)2)cos(x2)=2xcos2(x2).
Since the limits exist, the limit of this product is thus the product of the limits. And so we finally get
ddxtan(x2)=2xcos2(x2),
which some would write (using another trigonometric identity) as
ddxtan(x2)=2xsec2(x2).