The Fundamental Theorem of Calculus (Part-1) is an extremely powerful theorem that establishes the relationship between differentiation and integration.
If f(x) is continuous over an interval [a, b], and the function F(x) is defined by
Then the derivative of F(x) is
F'(x) = (x)'f(x) - (a)'f(a)
Using the First Fundamental Theorem of Integral Calculus, find the derivative of g(x) in each case.
Example 1 :
Solution :
Example 2 :
Solution :
Example 3 :
Solution :
Example 4 :
Solution :
g'(x) = (x^{3})'(cos x^{3}) - (1)'(cos 1)
g'(x) = (3x^{2})(cos x^{3}) - (0)(cos 1)
g'(x) = 3x^{2}cos x^{3}
Example 5 :
Solution :
g'(x) = (2x)'[(2x)^{3}] - (x)'(x^{3})
g'(x) = (2)(8x^{3}) - (1)(x^{3})
g'(x) = 16x^{3 }- x^{3}
g'(x) = 15x^{3}
Example 6 :
Solution :
g'(x) = (x^{3})'(cos x^{3}) - (x)'(cos x)
g'(x) = (3x^{2})(cos x^{3}) - (1)(cos x)
g'(x) = 3x^{2}cos x^{3} - cos x
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