05Class 12 Maths

Continuity and Differentiability formulas

Continuity conditions and all the differentiation rules for Class 12. This is the calculus you'll use for the rest of the year.

Continuity

Condition at a point
lim⁡x→c−f(x)=lim⁡x→c+f(x)=f(c)\displaystyle \lim_{x \to c^-}f(x) = \lim_{x \to c^+}f(x) = f(c)

Differentiation rules

Chain rule
dydx=dydu⋅dudx\displaystyle \frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx}
Parametric form
dydx=dy/dtdx/dt\displaystyle \frac{dy}{dx} = \frac{dy/dt}{dx/dt}
Logarithmic differentiation
y=uv  ⇒  dydx=uv[vududx+ln⁡u dvdx]\displaystyle y = u^v \;\Rightarrow\; \frac{dy}{dx} = u^v\left[\frac{v}{u}\frac{du}{dx} + \ln u\,\frac{dv}{dx}\right]

Standard derivatives

Exponential and log
ddxex=ex,ddxln⁡x=1x,ddxax=axln⁡a\displaystyle \frac{d}{dx}e^x = e^x, \quad \frac{d}{dx}\ln x = \frac1x, \quad \frac{d}{dx}a^x = a^x\ln a
Other trigonometric
ddxsec⁡x=sec⁡xtan⁡x,ddxcsc⁡x=−csc⁡xcot⁡x,ddxcot⁡x=−csc⁡2x\displaystyle \frac{d}{dx}\sec x = \sec x\tan x, \quad \frac{d}{dx}\csc x = -\csc x\cot x, \quad \frac{d}{dx}\cot x = -\csc^2x
Inverse sine and cosine
ddxsin⁡−1x=11−x2,ddxcos⁡−1x=−11−x2\displaystyle \frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1 - x^2}}, \quad \frac{d}{dx}\cos^{-1}x = -\frac{1}{\sqrt{1 - x^2}}
Inverse tangent
ddxtan⁡−1x=11+x2\displaystyle \frac{d}{dx}\tan^{-1}x = \frac{1}{1 + x^2}

Where students slip up

  • Assuming every continuous function is differentiable. |x| is continuous at 0 but isn't differentiable there.
  • Stopping the chain rule early with functions like sin(x²).
  • Using the power rule on xˣ. Take logs first.

Try it on your weakest chapter.

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