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Denition approach of learning new topics

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19<br />

(b) lim x→1 +<br />

(c) lim x→1 −<br />

(d) lim f(x)<br />

x→1<br />

(e) lim f(x)<br />

x→2 +<br />

(f) lim x→2 −<br />

(g) f(2)<br />

Solutions- a:2,b:1,c:1,d:1,e:0,f:-1,g:0<br />

Part III<br />

Lecture 3<br />

4 Derivatives<br />

<strong>Denition</strong>. Derivative <strong>of</strong> a function y = f(x) at a point x = a is dened as<br />

dy<br />

f(x) − f(a)<br />

dx∣ = lim<br />

x→a<br />

x=a<br />

x − a<br />

= lim<br />

h→0<br />

f(a + h) − f(a)<br />

h<br />

Above we found derivative <strong>of</strong> the function at some particular point x = a.<br />

What if we want to nd the derivative at any point x? We just replace a by x!<br />

Simple right?<br />

Notation : If given y = f(x),<br />

dy<br />

dx = d<br />

dx g(x) = g′ (x)<br />

So the derivative <strong>of</strong> a function y = f(x) at any point x is given as<br />

dy<br />

f(x + h) − f(x)<br />

dx∣ = lim<br />

at any x<br />

h→0 h<br />

Example. If f(x) = sin x, g(x) = x + 1<br />

x − 1<br />

then we have

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