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Rozwiązania różniczkowania: przewodnik krok po kroku

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Step-by-Step Solutions to Exercise 18.2 - Differentiation
1. Differentiate (a^2 - x^2)^10 w.r.t. x
Let y = (a^2 - x^2)^10
Using chain rule: dy/dx = 10(a^2 - x^2)^9 * d/dx (a^2 - x^2)
Since d/dx (a^2 - x^2) = -2x, we get: dy/dx = -20x (a^2 - x^2)^9
2. Differentiate log[log(log x)] w.r.t. x
Let y = log[log(log x)]
Using chain rule: dy/dx = 1/[log(log x)] * d/dx [log(log x)]
d/dx [log(log x)] = 1/log x * d/dx [log x] = 1/log x * (1/x)
Therefore, dy/dx = 1/[log(log x)] * (1/log x) * (1/x)
3. Differentiate cos(x^3) w.r.t. x
Let y = cos(x^3)
Using chain rule: dy/dx = -sin(x^3) * d/dx (x^3)
Since d/dx (x^3) = 3x^2, we get: dy/dx = -3x^2 sin(x^3)
4. Differentiate sin^3(sqrt x) w.r.t. x
Let y = sin^3(sqrt x) = [sin(sqrt x)]^3
Using chain rule: dy/dx = 3 [sin(sqrt x)]^2 * cos(sqrt x) * d/dx (sqrt x)
Since d/dx (sqrt x) = 1/(2 sqrt x), we get: dy/dx = 3 sin^2(sqrt x) * cos(sqrt x) * (1/(2 sqrt x))
5. Differentiate [log(cos x)]^2 w.r.t. x
Let y = [log(cos x)]^2
Using chain rule: dy/dx = 2[log(cos x)] * d/dx [log(cos x)]
d/dx [log(cos x)] = (1/cos x) * (-sin x) = -tan x
Therefore, dy/dx = 2 log(cos x) * (-tan x)
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