Keri464
Answered

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Find the derivative of the function.
y=sqrt( 3x+sqrt( 3x+sqrt( 3x))) Plz help :)

Sagot :

[tex]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][/tex]

[tex]=\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}}}} [/tex]
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