Skip to content
Question of 175

Q.Find the derivative of cos⁡x\cos x from first principle. OR Find the derivative of cos⁡x1+sin⁡x\dfrac{\cos x}{1+\sin x}.

Madhya Pradesh MpbseMP Board Higher Secondary (Class 11) 2023Subjective· 3mImportance★★★★★
0% · 0/175 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

From first principles, ddx(cos⁡x)=−sin⁡x\dfrac{d}{dx}(\cos x) = -\sin x.

By definition, f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \displaystyle\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}, with f(x)=cos⁡xf(x)=\cos x:

f′(x)=lim⁡h→0cos⁡(x+h)−cos⁡xhf'(x) = \displaystyle\lim_{h\to0}\dfrac{\cos(x+h)-\cos x}{h}

Using the identity cos⁡C−cos⁡D=−2sin⁡(C+D2)sin⁡(C−D2)\cos C - \cos D = -2\sin\left(\dfrac{C+D}{2}\right)\sin\left(\dfrac{C-D}{2}\right) with C=x+hC=x+h, D=xD=x:

cos⁡(x+h)−cos⁡x=−2sin⁡(x+h2)sin⁡(h2)\cos(x+h)-\cos x = -2\sin\left(x+\dfrac{h}{2}\right)\sin\left(\dfrac{h}{2}\right)

So:

…

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.