Разные примеры на нахождение производной
Задача 1. Найти производную: \(y = {x^{\sin {x^3}}}.\)
Решение.
\[y' = {x^{\sin {x^3}}}\left( {3\cos ({x^3}) \cdot {x^2} \cdot \ln x + \frac{{\sin {x^3}}}{x}} \right).\]
Задача 2. Найти производную.
\[y = \frac{{4x + 1}}{{16{x^2} + 8x + 3}} + \frac{1}{{\sqrt 2 }}arctg\frac{{4x + 1}}{{\sqrt 2 }}.\]
Решение.
\[\begin{array}{l}y' = \frac{{4(16{x^2} + 8x + 3) - (4x + 1)(32x + 8)}}{{{{(16{x^2} + 8x + 3)}^2}}} + \\ + \frac{1}{{\sqrt 2 }}\frac{1}{{1 + \frac{{{{(4x + 1)}^2}}}{2}}} \cdot \frac{4}{{\sqrt 2 }} = \\ = \frac{{ - 64{x^2} - 32x + 4}}{{{{(16{x^2} + 8x + 3)}^2}}} + \frac{4}{2} \cdot \frac{2}{{2 + {{(4x + 1)}^2}}} = \\ = \frac{{ - 64{x^2} - 32x + 4}}{{{{(16{x^2} + 8x + 3)}^2}}} + \frac{1}{{2 + 16{x^2} + 8x + 1}} = \\ = \frac{{ - 64{x^2} - 32x + 4 + 16{x^2} + 8x + 3}}{{{{(16{x^2} + 8x + 3)}^2}}} = \\ = \frac{{ - 48{x^2} - 24x + 7}}{{{{(16{x^2} + 8x + 3)}^2}}}.\end{array}\]
Задача 3. Найти производную.
\[y = 2\arcsin \frac{2}{{3x + 4}} + \sqrt {9{x^2} + 24x + 12} \]
Решение.
\[\begin{array}{l}y' = 2\frac{1}{{\sqrt {1 - \frac{4}{{{{(3x + 4)}^2}}}} }} \cdot \frac{{ - 2 \cdot 3}}{{{{(3x + 4)}^2}}} + \frac{{18x + 24}}{{2\sqrt {9{x^2} + 24x + 1{)^2}} }} = \\ = 2 \cdot \frac{{3x + 4}}{{\sqrt {9{x^2} + 24x + 16 - 4} }} \cdot \frac{{ - 6}}{{{{(3x + 4)}^2}}} + \\ + \frac{{ - 12}}{{(3x + 4)\sqrt {9{x^2} + 24x + 12} }} = \\ = \frac{{ - 12}}{{(3x + 4)\sqrt {9x{}^2 + 24x + 12} }} + \frac{{9x + 12}}{{\sqrt {9{x^2} + 24x + 12} }} = \\ = \frac{{ - 12 + (9x + 12)(3x + 4)}}{{(3x + 4)\sqrt {9{x^2} + 24x + 12} }} = \frac{{27{x^2} + 72x + 36}}{{(3x + 4)\sqrt {9{x^2} + 24x + 12} }}.\end{array}\]
Решение.
\[y' = {x^{\sin {x^3}}}\left( {3\cos ({x^3}) \cdot {x^2} \cdot \ln x + \frac{{\sin {x^3}}}{x}} \right).\]
Задача 2. Найти производную.
\[y = \frac{{4x + 1}}{{16{x^2} + 8x + 3}} + \frac{1}{{\sqrt 2 }}arctg\frac{{4x + 1}}{{\sqrt 2 }}.\]
Решение.
\[\begin{array}{l}y' = \frac{{4(16{x^2} + 8x + 3) - (4x + 1)(32x + 8)}}{{{{(16{x^2} + 8x + 3)}^2}}} + \\ + \frac{1}{{\sqrt 2 }}\frac{1}{{1 + \frac{{{{(4x + 1)}^2}}}{2}}} \cdot \frac{4}{{\sqrt 2 }} = \\ = \frac{{ - 64{x^2} - 32x + 4}}{{{{(16{x^2} + 8x + 3)}^2}}} + \frac{4}{2} \cdot \frac{2}{{2 + {{(4x + 1)}^2}}} = \\ = \frac{{ - 64{x^2} - 32x + 4}}{{{{(16{x^2} + 8x + 3)}^2}}} + \frac{1}{{2 + 16{x^2} + 8x + 1}} = \\ = \frac{{ - 64{x^2} - 32x + 4 + 16{x^2} + 8x + 3}}{{{{(16{x^2} + 8x + 3)}^2}}} = \\ = \frac{{ - 48{x^2} - 24x + 7}}{{{{(16{x^2} + 8x + 3)}^2}}}.\end{array}\]
Задача 3. Найти производную.
\[y = 2\arcsin \frac{2}{{3x + 4}} + \sqrt {9{x^2} + 24x + 12} \]
Решение.
\[\begin{array}{l}y' = 2\frac{1}{{\sqrt {1 - \frac{4}{{{{(3x + 4)}^2}}}} }} \cdot \frac{{ - 2 \cdot 3}}{{{{(3x + 4)}^2}}} + \frac{{18x + 24}}{{2\sqrt {9{x^2} + 24x + 1{)^2}} }} = \\ = 2 \cdot \frac{{3x + 4}}{{\sqrt {9{x^2} + 24x + 16 - 4} }} \cdot \frac{{ - 6}}{{{{(3x + 4)}^2}}} + \\ + \frac{{ - 12}}{{(3x + 4)\sqrt {9{x^2} + 24x + 12} }} = \\ = \frac{{ - 12}}{{(3x + 4)\sqrt {9x{}^2 + 24x + 12} }} + \frac{{9x + 12}}{{\sqrt {9{x^2} + 24x + 12} }} = \\ = \frac{{ - 12 + (9x + 12)(3x + 4)}}{{(3x + 4)\sqrt {9{x^2} + 24x + 12} }} = \frac{{27{x^2} + 72x + 36}}{{(3x + 4)\sqrt {9{x^2} + 24x + 12} }}.\end{array}\]
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