C3 differentiation help please

Given that y=secx+tanx, find dy/dx and show that it can be written in the form 1/1-sinx

My working out:

dy/dx = secxtanx+sec²x

then I converted them
= 1/cosx (√ sec²x -1 ) + 1/cos²x

Am I on the right track? If not please help mee! :tongue:

Reply 1

Blazy

Original post by queenfatso

Write the whole thing as one fraction:

\displaystyle \frac{dy}{dx} = \sec(x)\tan(x) + \sec^2(x) = \frac{\sin(x)+1}{\cos^2(x)}

. Rewrite the denominator in terms of a familiar identity and then think about difference of two squares.

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C3 differentiation help please

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