Estude! Com Prof. Caju

Complementando outro tópico que havia criado mais cedo. Também estou com problemas em resolver essa questão:

4) a) f(x) = ([tex3]7x^{2}[/tex3]/2*[tex3]\sqrt[5]{3x+1}[/tex3])+[tex3]\sqrt{3x+1}[/tex3]

Observe

Uma solução:

[tex3]f(x)=\frac{7x^2.\sqrt[5]{3x+1}}{2}+\sqrt{3x+1}[/tex3]

Derivando, vem;

[tex3]f'(x)=\left(\frac{7x^2.\sqrt[5]{3x+1}}{2}+\sqrt{3x+1}\right)'[/tex3]

[tex3]f'(x)=\frac{7}{2}.(x^2.\sqrt[5]{3x+1})'+(\sqrt{3x+1})'[/tex3]

[tex3]f'(x)=\frac{7}{2}.[
(x^2)'.\sqrt[5]{3x+1}+x^2.(\sqrt[5]{3x+1)'}]+[(3x+1)^\frac{1}{2}]'[/tex3]

[tex3]f'(x)=\frac{7}{2}.\{2x.\sqrt[5]{3x+1}+x^2.[(3x+1)^\frac{1}{5}]'\}+(3x+1)'.\frac{1}{2}(3x+1)^{-\frac{1}{2}}[/tex3]

[tex3]f'(x)=\frac{7}{2}.\left[2x.\sqrt[5]{3x+1}+x^2.(3x+1)'.\frac{1}{5}.(3x+1)^{-
\frac{4}{5}}\right]+\frac{3}{2\sqrt{3x+1}}[/tex3]

[tex3]f'(x)=\frac{7}{2}.\left[2x.\sqrt[5]{3x+1}+\frac{3x^2}{5(3x+1)^{\frac{4}{5}}}\right]+\frac{3}{2\sqrt{3x+1}}[/tex3]

[tex3]f'(x)=\frac{7}{2}.\left[2x.\sqrt[5]{3x+1}+\frac{3x^2}{5\sqrt[5]{(3x+1)^4}}\right]+\frac{3}{2\sqrt{3x+1}}[/tex3]

[tex3]f'(x)=\frac{7}{2}.\left[\frac{10x(3x+1)+3x^2}{5\sqrt[5]{(3x+1)^4}}\right]+\frac{3}{2\sqrt{3x+1}}[/tex3]

Logo,

[tex3]f'(x)=\frac{7}{2}.\left[\frac{33x^2+10x}{5\sqrt[5]{(3x+1)^4}}\right]+\frac{3}{2\sqrt{3x+1}}[/tex3]

Ou

[tex3]f'(x)=\frac{231x^2}{10\sqrt[5]{(3x+1)^4}}+\frac{7x}{\sqrt[5]{(3x+1)^4}}+\frac{3}{2\sqrt{3x+1}}[/tex3]

Ou

[tex3]f'(x)=\frac{231x^2+70x+15\sqrt[10]{(3x+1)^3}}{10\sqrt[5]{(3x+1)^4}}[/tex3]

Ou

Ah!!! Tem uma infinidade de representação dessa mesma resposta!

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Bons estudos!