Differentiate (a) (x + cosx)(x -tanx ).(b)x+cosx/secx

1.	Differentiate (a) (x + cosx)(x -tanx ).(b)x+cosx/secx
Answer» (a) Answer: $\frac{d}{dx} (x + cosx)(x - tanx)$ This is of the form uv : { $\frac{d}{dx} uv = u\frac{d}{dx}v + v\frac{d}{x} u$ } { u = (x + cosx) ; v = (x - tanx) } ∴ $(x+cosx)\frac{d}{dx} (x - tanx) + (x-tanx)\frac{d}{dx} (x+cosx)$ (Open BRACKETS and DIFF. each TERM) = (x + cosx) × (1 - sec²x) + (x - tanx) × (1 - sinx) = x + cosx - xsec²x - sec²x cosx + x - xsinx - tanx + tanx sinx 2x + cosx - xsec²x - secx - xsinx - tanx + tanx sinx (B) Answer: $\frac{d}{dx} (\frac{x+cosx}{secx})$ This is of the form u/v: ( $\frac{d}{dx}(\frac{u}{v}) = \frac{v\frac{d}{dx}u - u\frac{d}{dx}v }{v^{2} }$ ) ∴ $(secx)\frac{d}{dx} (x+cosx) - (x+cosx)\frac{d}{dx}(secx) / sec^{2}x$ (Open brackets and diff. each term) = (secx)(1 -sinx) - (x + cosx)(secx tanx) / (sec²x) = secx - sinx secx - x secx tanx - cosx secx tanx / sec²x = secx - tanx - x secx - tanx / sec²x

Answer»

(a) Answer:

$\frac{d}{dx} (x + cosx)(x - tanx)$