Question:
differentiate using first principle:- i) x^-3
Answer:
Given f(x) = (x)-3 = 1/x3
and f(x+h) = (x+h)-3 = 1/(x+h)3
Now using first principle,
dy/dx = limh->0 [{f(x+h) - f(x)}/h]
=> dy/dx = limh->0 [{1/(x+h)3 -1/x3 }/h]
=> dy/dx = limh->0 [{x3 - (x+h)3 }/{x3 *(x+h)3 *h}
=> dy/dx = limh->0 [{(x-x-h)*(x2 + (x+h)2 + x*(x+h)}/{x3 *(x+h)3 *h}]
=> dy/dx = limh->0 [{(-h)*x2 + (x+h)2 + x*(x+h)}/{x3 *(x+h)3 *h}]
=> dy/dx = - limh->0 [{x2 + (x+h)2 + x*(x+h)}/{x3 *(x+h)3 }]
=> dy/dx = - [{x2 + (x)2 + x*(x)}/{x3 *(x)3 }]
=> dy/dx = -3x2 /x6
=> dy/dx = -3/x4
=> dy/dx = -3x-4
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