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Let f(x)=2x+3sqrtx Find f'(x) when x=7. I'm having trouble with the conjugate. I ended up with -111/14, but isn't that wrong?
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Just differentiate it
Rewrite f(x) as f(x) = 2x + 3x^(1/2) So, f'(x) = 2+ (3/2)x^(-1/2) Therefore, f'(7) = 2 + (3/2)(1/sqrt(7))
Thanks, guys! I'm just really struggling with some of the basic algebra
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