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kiruha [24]
3 years ago
14

Suppose f(π/3) = 3 and f '(π/3) = −7,

Mathematics
1 answer:
insens350 [35]3 years ago
8 0
Et's give this a go:h(x) = cos(x) / f(x)
derivative (recall the quotient rule)h'(x) = [ f(x) * (-sin(x)) - cos(x)*f'(x) ] / [ f(x) ]^2
simplifyh'(x) = [ -sin(x)*f(x) - cox(x)*f '(x) ] / [ f(x) ]^2h'(π/3) = [ -sin(π/3)*f(π/3) - cox(π/3)*f '(π/3) ] / [ f(π/3) ]^2h'(π/3) = −(3–√/2)∗(3)−(1/2)∗(−7)/(3)2
h'(π/3) = (−33–√/2+7/2)/9

And you can further simplify if you want, I'll stop there.
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