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Question

Differentiate from first principle:

(v) sin(3x+1)

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Solution

Given:

f(x)=sin(3x+1)

The derivative of a function f(x) is defined as:

f(x)=limh0f(x+h)f(x)h

Putting f(x) in above expression, we get:

f(x)=limh0sin(3(x+h)+1)sin(3x+1)h

Rationalizing the numerator, we get:


f(x)=limh0sin(3(x+h)+1)sin(3x+1)h×sin(3x+3h+1)+sin(3x+1)sin(3x+3h+1)+sin(3x+1)

f(x)=limh0sin(3x+3h+1)sin(3x+1)h(sin(3x+3h+1)+sin(3x+1))

Applying the formula,

sincsind=2cos(c+d2)sin(cd2), we get


f(x)=limh02cos(3x+3h+1+3x+12)sin(3x+3h+13x12)h(sin(3x+3h+1)+sin(3x+1))

f(x)=limh02cos(6x+3h+22)sin3h2h(sin(3x+3h+1)+sin(3x+1))

f(x)=limh02cos(6x+3h+22)(sin(3x+3h+1)+sin(3x+1)).sin3h23h2.32

f(x)=3cos(6x+22)(sin(3x+1)+sin(3x+1))


[limh0sinhh=1]


f(x)=3cos(3x+1)2sin(3x+1)

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