Question

why d/dx tanx= sinx/cosx and d/dx tanx= sec^2x why both different

Answer 1

I'll show you trigonometric derivatives

Answer 2

http://www.sosmath.com/calculus/diff/der03/der03.html

Answer 3

only tanx= sinx/cosx where D/DX tanx=sec^2x

Answer 4

ryt??

Answer 5

noo d/dx sinx is not tanx
sinx/cosx=tanx :)

Answer 6

Remember these two derivatives:
\[\frac{d}{dx}\sin (x) = \cos(x)\] \[\frac{d}{dx}\cos (x) = -\sin (x)\]
and these two trig identities:
\[ \sec (x) = \frac{1}{\cos (x)} \]
\[ \sin^2 (x) + \cos^2 (x) = 1\]

Like you said,
\[\tan (x) = \frac{\sin (x)}{\cos (x)} \]
To take the derivative of \( \tan (x) \), you have to use the quotient rule:
\[\frac{d}{dx} \frac{f(x)}{g(x)} = \frac{g(x) f'(x) - f(x) g'(x)}{\left(g(x)\right)^2}\]
Applying the quotient rule to \(\tan (x)\):
\[\frac{d}{dx} \frac{\sin (x)}{\cos (x)} = \frac{\cos (x) \cos (x) - \sin (x) \left(- \sin (x) \right) }{\cos^2 (x) }\]
This simplifies to
\[\frac{d}{dx} \frac{\sin (x)}{\cos (x)} = \frac{\cos^2 (x) + \sin^2 (x) }{\cos^2 (x)}\]
Further simplification with the identities mentioned above will show that
\[\frac{d}{dx} \tan (x) = \sec^2 (x)\]