Question

value of d/dx(tanx) solve this
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Answer 1

Question :- find the value of d/dx (tan x) ?

Solution :-

→ f(x) = (tan x)

using tan x = sin x / cos x ,

→ f'(x) = (sin x / cos x)

using quotient rule now we get,

→ f'(x) = [d/dx (sin x) * cos x - sin x * d/dx( cos x)] / (cos x)²

→ f'(x) = [{cos x * cos x - sin x * (- sin x)}/ cos ²x]

→ f'(x) = (cos² x + sin ² x) / cos²x

using sin²A + cos²A = 1 in numerator,

→ f'(x) = (1/ cos ²x)

using (1/cosA) = sec A ,

→ f'(x) = sec² x (Ans.)

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Answer 2

Step-by-step explanation:

Given: tan x

To find: Derivative of tan x

Solution:

Step 1: Write tan x in terms of sin x and cos x

$tan \: x = \frac{sin \: x}{cos \: x}$

Step 2: Do differentiation using quotient rule

$\frac{d}{dx}\left( \frac{U}{V}\right) = \frac{V \frac{dU}{dx} - U \frac{dV}{dx} }{ {V}^{2} }$

here

U= sin x

V= cos x

$\frac{d}{dx} \left( \frac{sin \: x}{cos \: x} \right)= \frac{cos \: x \: \frac{d}{dx}(sin\:x) - sin \: x \: \frac{d}{dx}(cos \: x)}{ {cos}^{2}x }$

we know that

$\frac{d}{dx}(sin \: x) = cos \: x \\ \\ \frac{d}{dx}(cos \: x) = - sin \: x$

Thus,

$\frac{d}{dx} \left( \frac{sin \: x}{cos \: x} \right) = \frac{cos \: x. \: cos \: x - sin \: x \: ( - sin \: x) }{ {cos}^{2}x } \\ \\ \frac{d}{dx} \left( \frac{sin \: x}{cos \: x} \right) = \frac{ {cos}^{2} \: x + {sin}^{2} \: x \: }{ {cos}^{2}x }$

Step 3: Apply trigonometric identities

$\frac{d}{dx} \left( \frac{sin \: x}{cos \: x} \right) = \frac{1}{ {cos}^{2} x} \\ \\ \because \: {cos}^{2} x + {sin}^{2} x = 1 \\ \\ \frac{d}{dx}(tan \: x) = {sec}^{2} x \\ \\ \because \: \frac{1}{cos \: x} = sec \: x$

Final answer:

$\bold{\red{ \frac{d}{dx} (tan \: x) = {sec}^{2}x}}$

Hope it helps you.

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