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Question 1

Question 2

Решите задачу:

$1) \ y = \dfrac{x^{2} + 7x + 5}{x^{2} - 3x}$

$y' = \dfrac{(2x + 7)(x^{2} - 3x) - (x^{2} + 7x + 5)(2x - 3)}{(x^{2} - 3x)^{2}} =$

$= \dfrac{2x^{3} - 6x^{2} + 7x^{2} + 21x - 2x^{3} + 3x^{2} - 14x^{2} + 21x - 10x + 15}{(x^{2} - 3x)^{2}} =$

$= \dfrac{-10x^{2} - 10x + 15}{(x^{2} - 3x)^{2}}$

$2) \ y = \dfrac{x \ sin \ x}{1 + tg \ x} = \dfrac{x \ sin \ x}{1 + \dfrac{sin \ x}{cos \ x}} = \dfrac{x \ sin \ x}{\dfrac{cos \ x + sin \ x}{cos \ x}} = \dfrac{x \ sin \ x \ cos \ x}{cos \ x + sin \ x}$

$y' = \dfrac{((sin x + x cos x) cos x + x sin x (-sin x))(cos x + sin x) - x sin x cos x (-sin x + cos x)}{(cos x + sin x)^{2}}$

$= \dfrac{sin \ x \ cos ^{2}x + sin^{2}x \ cos \ x + x \ cos^{3}x - x \ sin^{3}x}{1 + sin \ 2x}$

$3) \ y = x \ sin \ x \ arctg \ x$

$y' = (sin \ x + x \ cos \ x) \ arctg \ x + x \ sin \ x \ \dfrac{1}{1 + x^{2}}$

$4) \ y = \dfrac{1}{\sqrt{x}} \bigg(e^{x^{2} - arctg \ x + 0,5 \ ln + 1} \bigg)x =$

$= \dfrac{1}{x^{\frac{1}{2}}} \ \cdotp e^{x^{2} - arctg \ x + \frac{1}{2} \ \cdotp ln + 1} \ x =\dfrac{1}{x^{\frac{1}{2}}} \ \cdotp e^{x^{2}} \ \cdotp e^{-arctg \ x} \ \cdotp e^{\frac{1}{2} \ \cdotp ln} \ \cdotp e^{1} x =$

$= \dfrac{1}{x^{\frac{1}{2}} \ \cdotp e^{x^{2}}} \ \cdotp e^{-arctg \ x} \ \cdotp ex = e^{x^{2}} \ \cdotp e^{-arctg \ x} \ \cdotp ex = e^{x^{2} - arctg \ x + 1} \ x$

$y' = e^{x^{2}- arctgx+1}\ \bigg(2x - \dfrac{1}{1 + x^{2}} \bigg) \ x + e^{x^{2} - arctgx + 1} =$

$= \dfrac{3e^{x^{2} - arctgx + 1} \ x^{2} + 2e^{x^{2} - arctgx + 1} \ x^{4} - e^{x^{2} - arctg x + 1} \ x + e^{x^{2} - arctgx+1}}{1 + x^{2}}$

$5) \ y = x^{\frac{1}{x}}$

$y' = e^{ln \ \cdotp \frac{1}{x}} \ \cdotp \bigg(\dfrac{1}{x} \ \cdotp \frac{1}{x} + ln \ \cdotp \bigg(-\dfrac{1}{x^{2}} \bigg) \bigg) = x^{\frac{1 - 2x}{x}} - x^{\frac{1 - 2x}{x}} \ \cdotp ln$