Ex 5.2, 1 - Differentiate sin (x2 + 5) - Chapter 5 Class 12

  1. Chapter 5 Class 12 Continuity and Differentiability (Term 1)
  2. Serial order wise

Transcript

Ex 5.2, 1 Differentiate the functions with respect to π‘₯ sin⁑(π‘₯2 + 5) y = sin (x2 + 5) We need to find derivative of 𝑦, 𝑀.π‘Ÿ.𝑑.π‘₯ (𝑑𝑦 )/𝑑π‘₯ = (𝑑(sin⁑(π‘₯2 + 5)))/𝑑π‘₯ = cos (π‘₯2 + 5) Γ— 𝑑(π‘₯2 + 5)/𝑑π‘₯ = cos (π‘₯2 + 5) Γ— ((𝑑(π‘₯2))/𝑑π‘₯+ (𝑑(5))/𝑑π‘₯) = cos (π‘₯2 + 5) Γ— (γ€–2π‘₯γ€—^(2βˆ’1) + 0) = cos (π‘₯2 + 5) Γ— 2π‘₯ = πŸπ’™ 𝒄𝒐𝒔⁑〖 (π’™πŸ + πŸ“)γ€—

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