Question

Differentiate (x+secx)(x-tanx)wrt x​

Answer 1

Answer:

f'(x) = x(2 - sec²x + secx tanx) - tanx + secx - sec³x + secx tan²x

Step-by-step explanation:

Let f(x) = (x+secx)(x-tanx)

⇒ f(x) = x² - x tanx + x secx - secx tanx

On differentiating both sides w.r.t. x, we get

$\tt{ f'(x) = \dfrac{d}{dx}(x^{2}) - \dfrac{d}{dx}(x\tan x) + \dfrac{d}{dx}(x\sec x) - \dfrac{d}{dx}(\sec x\tan x) }$

$\implies f'(x) = 2x - \{\,x\dfrac{d}{dx}(\tan x) + \tan x\dfrac{d}{dx}(x)\,\} + \{\,x\dfrac{d}{dx}(\sec x) + \sec x\dfrac{d}{dx}(x)\,\} - \{\,\sec x\dfrac{d}{dx}(\tan x) + \tan x\dfrac{d}{dx}(\sec x)\,\}$

⇒ f'(x) = 2x - (x.sec²x + tanx) + (x.secx.tanx + secx) - (secx.sec²x - secx.tanx.tanx)

⇒ f'(x) = 2x - x sec²x - tanx + x secx tanx + secx - sec³x + secx tan²x

⇒ f'(x) = x(2 - sec²x + secx tanx) - tanx + secx - sec³x + secx tan²x

Formula used :-

1) $\tt{ \dfrac{d}{dx}(I \times II) = I\dfrac{d}{dx}(II) + II\dfrac{d}{dx}(I) }$