If f(x) =sec^2(x/2) and g(x) = sqrt(5x-1 , what is f'(g(x)) ?

1 Answer
Oct 22, 2017

f'(g (x))=tan ((sqrt (5x-1))/2)xxsec^2 ((sqrt (5x-1))/2)

Explanation:

f'(x)=??
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f (x) " " is a composite function composed of two functions
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Trigonometric function u (x)=sec(x/2) and
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Polynomial function v (x)=x^2
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f'(x) is determined by applying the chain rule .
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f (x)=v (u (x))
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f'(x)=v'(u (x)) xx u'(x)
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v'(x)=(x^2)'=2x
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u'(x)=(sec(x/2))'=(x/2)'xxsec'(x/2)
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u'(x)=1/2xxtan(x/2)sec (x/2)
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So,f'(x) is determined by substituting the differentiated functions
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f'(x)=v'(u (x)) xx u'(x)=2 (u (x))xx1/2xxtan(x/2)sec (x/2)
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f'(x)=2sec (x/2)xx1/2xxtan(x/2)sec (x/2)
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f'(x)=tan (x/2)xxsec^2 (x/2)
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f'(g (x))=tan ((g (x))/2)xxsec^2 ((g (x))/2)
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f'(g (x))=tan ((sqrt (5x-1))/2)xxsec^2 ((sqrt (5x-1))/2)