Question

Пожалуйста помогите!

Answer 1

Ответ:

в объяснении

Объяснение:

1.

a)

 $\displaystyle \bigg(ln(x+1)-\frac{8}{x^4} +3x^3+6\sqrt[5]{x^3} \bigg)'=\\\\\\=\bigg(ln(x+1)\bigg)'-8(x^{-4})' +3(x^3)'+6(x^{3/5})'=\\\\\\= \frac{1}{x+1} -8*(-4*x^{-5})+3*3x^3+6*\frac{3}{5} x^{-2/5}=\\\\\\=\frac{1}{x+1} +\frac{32}{x^5} +9x^3+\frac{18}{5\sqrt[5]{x^2} }$

б)

$\displaystyle (uv)'=u'v+uv'\\\\\bigg(x^5*tg(4x+2)\bigg)=(x^5)'*tg(4x+2)+x^5(tg(4x+2))'=\\\\\\= 5x^4*tg(4x+2)+x^5*(tg(4x+2))'*(4x+2)'=\\\\\\=5x^4*tg(4x+2)+x^5*\frac{1}{cos^2(4x+2)} *4=5x^4tg(4x+2)+\frac{4x^5}{cos^2(4x+2)}$

в)

$\displaystyle \bigg(\frac{u}{v} \bigg)'=\frac{u'v-uv'}{v^2} \\\\\\\bigg(\frac{tg^3(2x)}{arctg(x^5)} \bigg)'=\frac{( tg^3(2x))'*arctg(x^5)-tg^3(2x)*(arctg(x^5))' }{arctg^2(x^5)} =\\\\\\=\frac{3tg^2(2x)*2*arctg(x^5)-tg^3(2x)* \displaystyle \frac{1}{1+(x^5)^2} *5x^4}{arctg^2(2x)} =\\\\\\=\frac{6tg^2(2x)}{arttg(x^5)} -\frac{tg^3(2x)*5x^4}{(1+x^{10})*arctg^2(2x)}$

2.

a)

$\displaystyle \lim_{x \to 5} \frac{2x^2-18x+40}{x^2-8x+15} =\frac{\displaystyle \lim_{x \to5} (2x^2-18x+40)' }{\displaystyle \lim_{x \to 5} (x^2-8x+15)' } =\frac{\displaystyle \lim_{x \to5} (4x-18) }{\displaystyle \lim_{x \to 5} (2x-8)} =\frac{2}{2} =1$

б)

$\displaystyle \lim_{x \to 0} \frac{7^{3x}-1}{5^{2x}-1} =\frac{\displaystyle \lim_{x \to 0} (7^{3x}-1)' }{\displaystyle \lim_{x \to 0} (5^{2x}-1)' } =\frac{\displaystyle \lim_{x \to 0} (3*7^{3x}*ln(7)) }{\displaystyle \lim_{x \to 0} (2*5^{2x}*ln(5)) }=\frac{3ln(7)}{2ln(5)}$

3.

уравнение касательной к функции f(x) в точке х₀

$\displaystyle y_k=f(x_0)+f'(x_0)(x-x_0)\\\\\\\underline {f(-1) = 3}\\\\\\f'(x) = \frac{1}{2} (2x^2-7x)^{1/2-1}(2x^2-7x)'=\frac{4x-7}{2\sqrt{2x^2-7x} }$

3.

$\displaystyle \underline { f'(-1) = -\frac{11}{6} }\\\\\\y_k = 3+\bigg(-\frac{11}{6} \bigg)*(x-(-1))\\\\\\y_k = 3-\frac{11}{6} (x+1)\\\\\\\underline {y^k=\frac{7}{6} -\frac{11}{6} x}$