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Chain Rule of Differentiation

lf fx=x+1ta...

Question

lf $f (x) = (x + 1) {tan}^{- 1} (e^{- 2 x})$ , then $f^{'} (0)$ is

\frac{π}{2} + 1

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\frac{π}{4} - 1

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\frac{π}{6} + 5

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\frac{π}{3} + 5

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Solution

The correct option is D $\frac{π}{4} - 1$
$f (x) = (x + 1) {tan}^{- 1} (e^{- 2 x}) \Rightarrow f^{'} (x) = {tan}^{- 1} (e^{- 2 x}) + \frac{(x + 1) . - 2 e^{- 2 x}}{1 + e^{- 4 x}} \Rightarrow f^{'} (0) = {tan}^{- 1} 1 + (- 1) = \frac{π}{4} - 1$

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Similar questions

f (x) = \frac{1 - tan x}{4 x - π}, x \neq \frac{π}{4}, x \in [0, \frac{π}{4}]

f (x)

[0, \frac{π}{2}]

f (\frac{π}{4})

f (x) = {sin}^{2} x + {sin}^{2} (x + \frac{π}{3}) + cos x cos (x + \frac{π}{3})

g (\frac{5}{4}) = 1

g (1) = 0

(g o f) (x) =

f (x) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} \frac{k cos x}{π - 2 x}, & x \neq \frac{π}{2} 3 & a t x = \frac{π}{2} \end{matrix}

x = \frac{π}{2}

k =

f'' (x) > 0, \forall x \in R, f' (3) = 0

g (x) = f ({tan}^{2} x - 2 tan x + 4), 0 < x < \frac{π}{2}

g (x)

f (x) = 2 s i n^{3} x - 3 s i n^{2} x + 12 s i n x + 5 \forall x \in (0, \frac{π}{2})

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