3 years ago

Need help with this please. Given the following functions: f(u)=tan(u) and g(x)=x^9. Find:

1 answer:

3 0

You have
.. f'(u) = sec(u)^2
Putting in u = g(x) = x^9, this becomes
.. f'(g(x)) = sec(x^9)^2

You know
(f(g(x))' = f'(g(x))*g'(x) . . . . . chain rule
From previous results, this is
.. sec(x^9)^2*9x^8

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Volgvan

Hello,

x² + 5x + 6 = 0

a = 1 ; b = 5 ; c = 6

∆ = b² - 4ac = 5² - 4 × 1 × 6 = 25 - 24 = 1 > 0

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$x_{1} = \frac{ - b - \sqrt{\Delta} }{2a} = \frac{ - 5 - 1}{2 \times 1} = \frac{ - 6}{2} = - 3$

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S = { -3 ; -2}

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1 year ago

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