![y=\sqrt{3x+\sqrt{3x+\sqrt{3x}}}\\\\y'=\left(\sqrt{3x+\sqrt{3x+\sqrt{3x}}}\right)'\\\\=\frac{1}{2\sqrt{3x+\sqrt{3x+\sqrt{3x}}}}\cdot\left[3+\frac{1}{2\sqrt{3x+\sqrt{3x}}}\cdot\left(3+\frac{1}{2\sqrt{3x}}\cdot3\right)\right]\\\\=\frac{1}{2\sqrt{3x+\sqrt{3x+\sqrt{3x}}}}\cdot\left[3+\frac{1}{2\sqrt{3x+\sqrt{3x}}}\cdot\frac{6\sqrt{3x}+3}{2\sqrt{3x}}\right]](/tpl/images/0202/0126/23a81.png)
![=\frac{1}{2\sqrt{3x+\sqrt{3x+\sqrt{3x}}}}\cdot\left[3+\frac{6\sqrt{3x}+3}{4\sqrt{9x^2+3x\sqrt{3x}}}\right]\\\\=\frac{1}{2\sqrt{3x+\sqrt{3x+\sqrt{3x}}}}\cdot\frac{12\sqrt{9x^2+3x\sqrt{3x}}+6\sqrt{3x}+3}{4\sqrt{9x^2+3x\sqrt{3x}}}\\\\=\frac{12\sqrt{9x^2+3x\sqrt{3x}}+6\sqrt{3x}+3}{8\sqrt{27x^3+9x^2\sqrt{3x}+9x^2\sqrt{3x+\sqrt{3x}}+3x\sqrt{9x^2+3x\sqrt{3x}}}}](/tpl/images/0202/0126/5c62d.png)
Find the derivative of the function.
y=sqrt( 3x+sqrt( 3x+sqrt( plz : )...
Mathematics, 27.08.2019 09:20 whohelpme
Find the derivative of the function.
y=sqrt( 3x+sqrt( 3x+sqrt( plz : )
Answers: 1
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