
Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Last updated at May 29, 2023 by Teachoo
Ex 5.4, 4 Differentiate π€.π.π‘. π₯ in , sinβ‘γ (tan^(β1) π^(βπ₯) )γLet π¦ = sinβ‘γ (tan^(β1) π^(βπ₯) )γ Differentiating both sides π€.π.π‘.π₯ π¦^β² = (sinβ‘(tan^(β1) π^(βπ₯) ) )^β² = γcos γβ‘(tan^(β1) π^(βπ₯) ) Γ (tan^(β1) π^(βπ₯) )^β² = γcos γβ‘(tan^(β1) π^(βπ₯) ) Γ 1/(1 + (π^(βπ₯) )^2 ) Γ(π^(βπ₯) )^β² = γcos γβ‘(tan^(β1) π^(βπ₯) ) Γ 1/(1 + (π^(βπ₯) )^2 ) Γ βπ^(βπ₯) = (π^(βπ₯) γcos γβ‘(tan^(β1) π^(βπ₯) ))/(1 + (π^(βπ₯) )^2 ) = (βπ^(βπ) γππ¨π¬ γβ‘(γπππγ^(βπ) π^(βπ) ))/(π + π^(βππ) ) = (π^(βπ₯) γcos γβ‘(tan^(β1) π^(βπ₯) ))/(1 + (π^(βπ₯) )^2 ) = (βπ^(βπ) γππ¨π¬ γβ‘(γπππγ^(βπ) π^(βπ) ))/(π + π^(βππ) ) = (π^(βπ₯) γcos γβ‘(tan^(β1) π^(βπ₯) ))/(1 + (π^(βπ₯) )^2 ) = (βπ^(βπ) γππ¨π¬ γβ‘(γπππγ^(βπ) π^(βπ) ))/(π + π^(βππ) )