How do you differentiate $(2x+1)(x-tanx)$ ?

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Answer 1 · 2016-06-15T12:51:10

$d/(dx)(2x+1)(x-tanx)$

= $4x+1-2xsec^2x-sec^2x-2tanx$

If $f(x)=g(x)xxh(x)$ , $(df)/(dx)=g(x)xx(dh)/(dx)+(dg)/(dx)xxh(x)$

Hence $d/(dx)(2x+1)(x-tanx)$

= $(2x+1)(1-sec^2x)+2(x-tanx)$

= $2x+1-2xsec^2x-sec^2x+2x-2tanx$

= $4x+1-2xsec^2x-sec^2x-2tanx$

How do you differentiate (2x+1)(x-tanx) (2x+1)(x-tanx)?