Answer:
[tex]\boxed{\sf \: \:\dfrac{d}{dx}\dfrac{tanx - cotx}{tanx + cotx} = 2 \: sin2x \: }\\ \\ [/tex]
Step-by-step explanation:
[tex]\sf \:\dfrac{d}{dx}\dfrac{tanx - cotx}{tanx + cotx} \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{(tanx + cotx)\dfrac{d}{dx}(tanx - cotx) - (tanx - cotx)\dfrac{d}{dx}(tanx + cotx)}{ {(tanx + cotx)}^{2} } \\ \\ [/tex]
[tex]\sf \: = \dfrac{(tanx + cotx)( {sec}^{2}x + {cosec}^{2}x) - (tanx - cotx)x( {sec}^{2}x - {cosec}^{2}x) }{ {(tanx + cotx)}^{2} } \\ \\ [/tex]
[tex]\sf \: = \dfrac{(tanx + cotx)( 1 + {tan}^{2}x +1 + {cot}^{2}x) - (tanx - cotx)x(1 + {tan}^{2}x - 1 - {cot}^{2}x) }{ {(tanx + cotx)}^{2} } \\ \\ [/tex]
[tex]\sf \: = \dfrac{(tanx + cotx)({tan}^{2}x+ {cot}^{2}x + 2) - (tanx - cotx)x({tan}^{2}x - {cot}^{2}x) }{ {(tanx + cotx)}^{2} } \\ \\ [/tex]
[tex]\sf \: = \dfrac{(tanx + cotx) {(tanx + cotx)}^{2} - (tanx - cotx)^{2}(tanx + cotx) }{ {(tanx + cotx)}^{2} } \\ \\ [/tex]
[tex]\sf \: = \dfrac{(tanx + cotx) [ {(tanx + cotx)}^{2} - (tanx - cotx)^{2}]}{ {(tanx + cotx)}^{2} } \\ \\ [/tex]
[tex]\sf \: = \dfrac{4tanx \: cotx}{ tanx + cotx } \\ \\ [/tex]
[tex]\sf \: = \dfrac{4}{ tanx + \dfrac{1}{tanx} } \\ \\ [/tex]
[tex]\sf \: = \dfrac{4}{ \dfrac{ {tan}^{2}x + 1}{tanx} } \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{4tanx}{1 + {tan}^{2} x} \\ \\ [/tex]
[tex]\sf \: = \: 2 \times \dfrac{2tanx}{1 + {tan}^{2} x} \\ \\ [/tex]
[tex]\sf \: = \: 2 \: sin2x\\ \\ [/tex]
Hence,
[tex]\implies\sf \: \sf \:\dfrac{d}{dx}\dfrac{tanx - cotx}{tanx + cotx} = 2 \: sin2x \\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Formulae Used:
[tex]\sf \: \dfrac{d}{dx}\left(\dfrac{u}{v} \right) = \dfrac{v\dfrac{d}{dx}u - u\dfrac{d}{dx}v}{ {v}^{2} } \\ \\ [/tex]
[tex]\sf \: \dfrac{d}{dx}tanx = {sec}^{2}x \\ \\ [/tex]
[tex]\sf \: \dfrac{d}{dx}cotx = - {cosec}^{2}x \\ \\ [/tex]
[tex]\sf \: {(x + y)}^{2} - {(x - y)}^{2} = 4xy \\ \\ [/tex]
[tex]\sf \: tanx \times cotx = 1 \\ \\ [/tex]
[tex]\sf \: sin2x = \frac{2tanx}{1 + {tan}^{2} x} \\ \\ [/tex]
2 sin 2x
Formula: d/dx (u/v) = (v.du/dx - u.dv/dx) / v²
d/dx (tan x - cot x / tanx + cot x)
= [ (tan x + cot x) d ( tanx - cot x )/dx - (tan x - cot x) d ( tan x - cot x )/dx ] / (tan x + cot x)²
= [ ( tan x + cot x ) ( sec²x + cosec²x ) - ( tan x - cot x ) ( sec²x - cosec² x) ] / (tan x + cot x)² [as, d(tan x)/dx = sec x , d(cot x)/dx = -cosec² x ]
= 2 [ ( tan x cosec² x ) ( cot x sec² x) ] / (tan x + cot x)²
= 2 [ 2 / ( sin x.cos x) ] / [ sin² x + cos² x / sinx.cosx ]² [ as, cosec x = 1/sin x , sec x = 1/cos x ]
= [ 4 / sin x.cos x ] / [ 1 / sin²x.cos² x ] [ as, sin² x + cos² x = 1 ]
= 4 sin x . cos x
= 2 ( 2 sin x . cos x )
= 2 sin 2x [ as, 2sin x cos x = sin 2x ]
Hence, the differentiation of (tan x - cot x / tanx + cot x) w.r.t. x is 2 sin 2x ( ans. )
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Verified answer
Answer:
[tex]\boxed{\sf \: \:\dfrac{d}{dx}\dfrac{tanx - cotx}{tanx + cotx} = 2 \: sin2x \: }\\ \\ [/tex]
Step-by-step explanation:
[tex]\sf \:\dfrac{d}{dx}\dfrac{tanx - cotx}{tanx + cotx} \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{(tanx + cotx)\dfrac{d}{dx}(tanx - cotx) - (tanx - cotx)\dfrac{d}{dx}(tanx + cotx)}{ {(tanx + cotx)}^{2} } \\ \\ [/tex]
[tex]\sf \: = \dfrac{(tanx + cotx)( {sec}^{2}x + {cosec}^{2}x) - (tanx - cotx)x( {sec}^{2}x - {cosec}^{2}x) }{ {(tanx + cotx)}^{2} } \\ \\ [/tex]
[tex]\sf \: = \dfrac{(tanx + cotx)( 1 + {tan}^{2}x +1 + {cot}^{2}x) - (tanx - cotx)x(1 + {tan}^{2}x - 1 - {cot}^{2}x) }{ {(tanx + cotx)}^{2} } \\ \\ [/tex]
[tex]\sf \: = \dfrac{(tanx + cotx)({tan}^{2}x+ {cot}^{2}x + 2) - (tanx - cotx)x({tan}^{2}x - {cot}^{2}x) }{ {(tanx + cotx)}^{2} } \\ \\ [/tex]
[tex]\sf \: = \dfrac{(tanx + cotx) {(tanx + cotx)}^{2} - (tanx - cotx)^{2}(tanx + cotx) }{ {(tanx + cotx)}^{2} } \\ \\ [/tex]
[tex]\sf \: = \dfrac{(tanx + cotx) [ {(tanx + cotx)}^{2} - (tanx - cotx)^{2}]}{ {(tanx + cotx)}^{2} } \\ \\ [/tex]
[tex]\sf \: = \dfrac{4tanx \: cotx}{ tanx + cotx } \\ \\ [/tex]
[tex]\sf \: = \dfrac{4}{ tanx + \dfrac{1}{tanx} } \\ \\ [/tex]
[tex]\sf \: = \dfrac{4}{ \dfrac{ {tan}^{2}x + 1}{tanx} } \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{4tanx}{1 + {tan}^{2} x} \\ \\ [/tex]
[tex]\sf \: = \: 2 \times \dfrac{2tanx}{1 + {tan}^{2} x} \\ \\ [/tex]
[tex]\sf \: = \: 2 \: sin2x\\ \\ [/tex]
Hence,
[tex]\implies\sf \: \sf \:\dfrac{d}{dx}\dfrac{tanx - cotx}{tanx + cotx} = 2 \: sin2x \\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Formulae Used:
[tex]\sf \: \dfrac{d}{dx}\left(\dfrac{u}{v} \right) = \dfrac{v\dfrac{d}{dx}u - u\dfrac{d}{dx}v}{ {v}^{2} } \\ \\ [/tex]
[tex]\sf \: \dfrac{d}{dx}tanx = {sec}^{2}x \\ \\ [/tex]
[tex]\sf \: \dfrac{d}{dx}cotx = - {cosec}^{2}x \\ \\ [/tex]
[tex]\sf \: {(x + y)}^{2} - {(x - y)}^{2} = 4xy \\ \\ [/tex]
[tex]\sf \: tanx \times cotx = 1 \\ \\ [/tex]
[tex]\sf \: sin2x = \frac{2tanx}{1 + {tan}^{2} x} \\ \\ [/tex]
Answer:
2 sin 2x
Step-by-step explanation:
Formula: d/dx (u/v) = (v.du/dx - u.dv/dx) / v²
d/dx (tan x - cot x / tanx + cot x)
= [ (tan x + cot x) d ( tanx - cot x )/dx - (tan x - cot x) d ( tan x - cot x )/dx ] / (tan x + cot x)²
= [ ( tan x + cot x ) ( sec²x + cosec²x ) - ( tan x - cot x ) ( sec²x - cosec² x) ] / (tan x + cot x)² [as, d(tan x)/dx = sec x , d(cot x)/dx = -cosec² x ]
= 2 [ ( tan x cosec² x ) ( cot x sec² x) ] / (tan x + cot x)²
= 2 [ 2 / ( sin x.cos x) ] / [ sin² x + cos² x / sinx.cosx ]² [ as, cosec x = 1/sin x , sec x = 1/cos x ]
= [ 4 / sin x.cos x ] / [ 1 / sin²x.cos² x ] [ as, sin² x + cos² x = 1 ]
= 4 sin x . cos x
= 2 ( 2 sin x . cos x )
= 2 sin 2x [ as, 2sin x cos x = sin 2x ]
Hence, the differentiation of (tan x - cot x / tanx + cot x) w.r.t. x is 2 sin 2x ( ans. )