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What is the differential of the function y equals the square root of the 3x-5?
Wiki User
∙ 11y ago
(f(g(t)))'=f'(g(t)) x g'(t)
y=sqrt(3x-5) here f(t)=sqrt(t) and g(t)=3t-5, f'(t)=1/(2 sqrt(t)) and g'(t) = 3
So y'= f'(g(x)) x g'(x) = 1/(2 sqrt(3x-5) x 3 = 3/(2 sqrt(3x-5))
Wiki User
∙ 11y ago
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