Examples

Chapter 5 Class 12 Continuity and Differentiability
Serial order wise

### Transcript

Example 40 (Method 1) Differentiate the following w.r.t. x. (ii) tan β1 (sinβ‘π₯/( 1 +γ cosγβ‘γπ₯ γ )) Let π(π₯) = tan β1 (πππβ‘π/( 1 +γ πππγβ‘γπ γ )) π(π₯) = tan β1 ((π γπ¬π’π§ γβ‘γπ/πγ γ ππ¨π¬ γβ‘γπ/πγ)/( 1+ (π γππ¨π¬γ^πβ‘γ π/πγ β π))) = tan β1 ((2 γsin γβ‘γπ₯/2γ γ cos γβ‘γπ₯/2 γ)/( 2 cos^2β‘γ π₯/2γ )) = tan β1 ((2 γsin γβ‘γπ₯/2γ)/( 2 cosβ‘γπ₯/2γ )) We know that sin 2ΞΈ = 2 sin ΞΈ cos ΞΈ Replacing ΞΈ by π/2 sin ΞΈ = 2 πππβ‘γπ½/πγ πππβ‘γπ½/πγ and cos 2ΞΈ = 2cos2 ΞΈ β 1 Replacing ΞΈ by π/2 cos ΞΈ = 2cos2 π½/π β 1 = tan β1 (γsin γβ‘γπ₯/2γ/( cosβ‘γ π₯/2γ )) = tan β1 (γtan γβ‘γπ₯/2γ ) = π/π π(π) = π/π Differentiating π€.π.π‘.π₯ πβ(π₯) = 1/2 (π(π₯))/ππ₯ πβ(π) = π/π (As tan^(β1)β‘γ(tanβ‘π)γ =π) Example 40 (Method 2) Differentiate the following w.r.t. x. (ii) tan β1 (sinβ‘π₯/( 1 +γ cosγβ‘γπ₯ γ )) Let π(π₯) = tan β1 (πππβ‘π/( 1 +γ πππγβ‘γπ γ )) Differentiating w.r.t x π^β² (π₯) = 1/(1 + (πππβ‘π/( 1 +γ πππγβ‘γπ γ ))^2 ) (πππβ‘π/( 1 +γ πππγβ‘γπ γ ))^β² = 1/(((1 + cosβ‘π₯ )^2 + sin^2β‘π₯)/(1 + cosβ‘π₯ )^2 ) (((πππβ‘π )^β² (π + πππ π)β(π + πππ π)^β² πππ π)/( (1 +γ πππγβ‘π )^π )) = (1 + cosβ‘π₯ )^2/((1 + cosβ‘π₯ )^2 + sin^2β‘π₯ ) ((πππ π(π + πππ π). β (β πππ π)πππ π)/( (1 +γ πππγβ‘π )^π ))= (πππ  π₯ + πππ^π π + πππ^π π)/(1 + γππ¨π¬γ^πβ‘π + 2 cosβ‘π₯ + γπππγ^πβ‘π ) = (πππ π +π)/(1 + 1 + 2 cosβ‘π₯ ) = (πππ π + π)/(2 + 2 cosβ‘π₯ ) = (1 + πππ π)/(2(1 + cosβ‘π₯) ) = 1/2