Finding derivative of Inverse trigonometric functions

Chapter 5 Class 12 Continuity and Differentiability
Concept wise

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Example 27 Find the derivative of f given by f (x) = tanβ1 π₯ assuming it exists. π (π₯)=γπ‘ππγ^(β1) π₯ Let π= γπππγ^(βπ) π tanβ‘γπ¦=π₯γ π=π­ππ§β‘γπ γ Differentiating both sides π€.π.π‘.π₯ ππ₯/ππ₯ = (π (tanβ‘π¦ ))/ππ₯ 1 = (π (tanβ‘π¦ ))/ππ₯ Γ ππ¦/ππ¦ 1 = (π (tanβ‘π¦ ))/ππ¦ Γ ππ¦/ππ₯ 1 = γπ¬ππγ^π π . ππ¦/ππ₯ 1 = (π + πππππ) ππ¦/ππ₯ ππ¦/ππ₯ = 1/(1 + γπ­ππ§γ^πβ‘π ) Putting π‘ππβ‘π¦ = π₯ ππ¦/ππ₯ = 1/(1 + π^π ) Hence (π(γπ­ππ§γ^(βπ)β‘γπ)γ)/ππ = π/(π + π^π ) As π¦ = γπ‘ππγ^(β1) π₯ So, πππβ‘π = π Derivative of γπππγ^(βπ) π π (π₯)=γπππ γ^(β1) π₯ Let π= γπππγ^(βπ) π cosβ‘γπ¦=π₯γ π=ππ¨π¬β‘γπ γ Differentiating both sides π€.π.π‘.π₯ ππ₯/ππ₯ = (π (cosβ‘π¦ ))/ππ₯ 1 = (π (cosβ‘π¦ ))/ππ₯ Γ ππ¦/ππ¦ 1 = (π (cosβ‘π¦ ))/ππ¦ Γ ππ¦/ππ₯ 1 = (βsinβ‘π¦) ππ¦/ππ₯ (β1)/sinβ‘π¦ =ππ¦/ππ₯ ππ¦/ππ₯ = (β1)/πππβ‘π ππ¦/ππ₯= (β1)/β(π β γπππγ^π π) Putting πππ β‘γπ¦=π₯γ ππ¦/ππ₯= (β1)/β(1 β π^π ) Hence, (π(γπππγ^(βπ) π" " ))/ππ = (βπ)/β(π β π^π ) "We know that" γπ ππγ^2 π+γπππ γ^2 π=1 γπ ππγ^2 π=1βγπππ γ^2 π πππβ‘π½=β(πβγπππγ^π π½) " " As π¦ = γπππ γ^(β1) π₯ So, πππβ‘π = π Derivative of γπππγ^(βπ) π π (π₯)=γπππ‘γ^(β1) π₯ Let π= γπππγ^(βπ) π cotβ‘γπ¦=π₯γ π=ππ¨π­β‘γπ γ Differentiating both sides π€.π.π‘.π₯ ππ₯/ππ₯ = (π (cotβ‘π¦ ))/ππ₯ 1 = (π (cotβ‘π¦ ))/ππ₯ Γ ππ¦/ππ¦ 1 = (π (cotβ‘π¦ ))/ππ¦ Γ ππ¦/ππ₯ 1 = βππ¨γπ¬ππγ^π π . ππ¦/ππ₯ 1 = β(π +πππππ) ππ¦/ππ₯ ππ¦/ππ₯ = (β1)/(1 + γππ¨π­γ^πβ‘π ) Putting πππ‘β‘π¦ = π₯ ππ¦/ππ₯ = (β1)/(π^π + π) Hence (π(γππ¨π­γ^(βπ)β‘γπ)γ)/ππ = (βπ)/(π^π + π) (π΄π  γ πππ ππγ^2β‘γπ¦= γ1+γβ‘γπππ‘γ^2β‘π¦ γ) As π¦ = γπππ‘γ^(β1) π₯ So, πππβ‘π = π Derivative of γπππγ^(βπ) π π (π₯)=γπ ππγ^(β1) π₯ Let π= γπππγ^(βπ) π secβ‘γπ¦=π₯γ π=π¬ππβ‘γπ γ Differentiating both sides π€.π.π‘.π₯ ππ₯/ππ₯ = (π (secβ‘π¦ ))/ππ₯ 1 = (π (secβ‘π¦ ))/ππ₯ Γ ππ¦/ππ¦ 1 = (π (secβ‘π¦ ))/ππ¦ Γ ππ¦/ππ₯ 1 = πππβ‘π .πππβ‘π. ππ¦/ππ₯ ππ¦/ππ₯ = 1/(πππβ‘π .γ secγβ‘π¦ ) ππ¦/ππ₯ = 1/((β(γπ¬ππγ^πβ‘π β π)) .γ secγβ‘π¦ ) Putting value of π ππβ‘π¦ = π₯ ππ¦/ππ₯ = 1/((β(π₯^2 β 1 ) ) . π₯) ππ¦/ππ₯ = 1/(π₯ β(π₯^2 β 1 ) ) Hence π(γπππγ^(βπ) π)/ππ = π/(π β(π^π β π ) ) As tan2 ΞΈ = sec2 ΞΈ β 1, tan ΞΈ = β("sec2 ΞΈ β 1" ) As π¦ = γπ ππγ^(β1) π₯ So, πππβ‘π = πDerivative of γπππππγ^(βπ) π π (π₯)=γπππ ππγ^(β1) π₯ Let π= γπππππγ^(βπ) π cosecβ‘γπ¦=π₯γ π=ππ¨π¬ππβ‘γπ γ 1 = (π (cosecβ‘π¦ ))/ππ¦ Γ ππ¦/ππ₯ 1 = βcosecβ‘π¦ .cotβ‘π¦ . ππ¦/ππ₯ ππ¦/ππ₯ = 1/(γβcosecγβ‘π¦ .πππβ‘π ) ππ¦/ππ₯ = 1/(γβcosecγβ‘π¦ . β(γππ¨πππγ^πβ‘π β π)) Putting value of πππ ππβ‘π¦ = π₯ ππ¦/ππ₯ = (β1)/(π₯ β(π₯^2 β 1 ) ) Hence π(γπππππγ^(βπ) π)/ππ = (βπ)/(π β(π^π β π ) ) As cot2 ΞΈ = cosec2 ΞΈ β 1, cot ΞΈ = β("cosec2 ΞΈ β 1" ) Differentiating both sides π€.π.π‘.π₯ ππ₯/ππ₯ = (π (cosecβ‘π¦ ))/ππ₯ 1 = (π (cosecβ‘π¦ ))/ππ₯ Γ ππ¦/ππ¦ As cot2 ΞΈ = cosec2 ΞΈ β 1, cot ΞΈ = β("cosec2 ΞΈ β 1" ) As π¦ = coγπ ππγ^(β1) π₯ So, coπππβ‘π = π