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NCERT Exemplar · Q31

Q.Differentiate w.r.t. xx: cos⁡(tan⁡x+1)\cos\left(\tan\sqrt{x + 1}\right).

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We differentiate cos⁡(tan⁡x+1)\cos(\tan\sqrt{x+1}) by applying the Chain Rule three times in succession. The final derivative is −sin⁡(tan⁡x+1)⋅sec⁡2(x+1)⋅12x+1-\sin(\tan\sqrt{x+1}) \cdot \sec^2(\sqrt{x+1}) \cdot \frac{1}{2\sqrt{x+1}}.

The problem asks us to differentiate a composition of three functions: cosine, tangent, and square root. When you have a function inside a function inside another function, the Chain Rule is your only tool — and it works exactly the same way as for two functions, just applied repeatedly.

Think of it like peeling an onion: start from the outermost layer and work inward. Each layer you peel gives you a derivative factor, and you multiply them all together.

Let’s name the layers clearly:

  • Outermost: cos⁡(something)\cos(\text{something})
  • Middle: tan⁡(something)\tan(\text{something})
  • Innermost: x+1\sqrt{x+1}

We’ll differentiate step by step.


  1. Differentiate the outermost function. The derivative of cos⁡(u)\cos(u) with respect to uu is −sin⁡(u)-\sin(u). Here u=tan⁡x+1u = \tan\sqrt{x+1}. So the first factor is:

−sin⁡(tan⁡x+1)-\sin\left(\tan\sqrt{x+1}\right)

  1. Multiply by the derivative of the middle function. Now we need the derivative of tan⁡(v)\tan(v) with respect to vv, where v=x+1v = \sqrt{x+1}. The derivative of tan⁡v\tan v is sec⁡2v\sec^2 v. So the second factor is:

sec⁡2(x+1)\sec^2\left(\sqrt{x+1}\right)

  1. Multiply by the derivative of the innermost function. Finally, differentiate x+1\sqrt{x+1} with respect to xx. Write it as (x+1)1/2(x+1)^{1/2}. Its derivative is:

12(x+1)−1/2=12x+1\frac{1}{2}(x+1)^{-1/2} = \frac{1}{2\sqrt{x+1}}

  1. Multiply all three factors together. The complete derivative is: …

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