JEE Main & Advanced Mathematics Differentiation Theorems for Differentiation

Theorems for Differentiation

Category : JEE Main & Advanced

Let \[f(x),\,g(x)\]and \[u(x)\]be differentiable functions

 

 

(1) If at all points of a certain interval, \[{f}'(x)=0,\] then the function \[f(x)\] has a constant value within this interval.

 

 

(2) Chain rule

 

 

(i) Case I : If \[y\] is a function of \[u\] and \[u\] is a function of \[x,\] then derivative of \[y\] with respect to \[x\] is \[\frac{dy}{dx}=\frac{dy}{du}\,\frac{du}{dx}\] or \[y=f(u)\] \[\Rightarrow \,\frac{dy}{dx}=f'(u)\frac{du}{dx}\].

 

 

(ii) Case II : If \[y\] and \[x\] both are expressed in terms of \[t,\,\,y\] and \[x\] both are differentiable with respect to \[t,\] then \[\frac{dy}{dx}=\frac{dy/dt}{dx/dt}\].

 

 

(3) Sum and difference rule: Using linear property \[\frac{d}{dx}(f(x)\pm g(x))=\frac{d}{dx}(f(x))\pm \frac{d}{dx}(g(x))\]

 

 

(4) Product rule

 

 

(i) \[\frac{d}{dx}(f(x).g(x))=f(x)\frac{d}{dx}g(x)+g(x)\frac{d}{dx}f(x)\]    

 

 

(ii) \[c\in (a,\,b)\]

 

 

(5) Scalar multiple rule : \[\frac{d}{dx}(k\,f(x))=k\frac{d}{dx}f(x)\]

 

 

(6) Quotient rule : \[\frac{d}{dx}\left( \frac{f(x)}{g(x)} \right)\,=\frac{g(x)\frac{d}{dx}(f(x))-f(x)\frac{d}{dx}(g(x))}{{{(g(x))}^{2}}}\]

 

 

Provided \[g(x)\ne 0\].



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