Respuesta :
Given:
[tex]y=3x\sqrt{x^4-5}[/tex]Required:
We need to differentiate the given expression.
Explanation:
Consider the given expression.
[tex]y=3x\sqrt{x^4-5}[/tex][tex]y=3x(x^4-5)^{\frac{1}{2}}[/tex]Differentiate the given expression with respect to x.
[tex]Use\text{ }(uv)^{\prime}=uv^{\prime}+vu^{\prime}.\text{ Here u=3x and v=}(x^4-5)^{\frac{1}{2}}.[/tex][tex]y^{\prime}=3x(\frac{1}{2})(x^4-5)^{\frac{1}{2}-1}(4x^3)+(x^4-5)^{\frac{1}{2}}(3)[/tex][tex]y^{\prime}=\frac{3x(4x^3)}{2\left(x^4-5\right)^{\frac{1}{2}}}+3(x^4-5)^{\frac{1}{2}}[/tex][tex]y^{\prime}=\frac{6x^4}{\left(x^4-5\right)^{\frac{1}{2}}}+3(x^4-5)^{\frac{1}{2}}[/tex][tex]y^{\prime}=\frac{6x^4}{\left(x^4-5\right)^{\frac{1}{2}}}+\frac{3(x^4-5)^{\frac{1}{2}}(x^4-5)^{\frac{1}{2}}}{(x^4-5)^{\frac{1}{2}}}[/tex][tex]y^{\prime}=\frac{6x^4}{\left(x^4-5\right)^{\frac{1}{2}}}+\frac{3(x^4-5)}{(x^4-5)^{\frac{1}{2}}}[/tex][tex]y^{\prime}=\frac{6x^4}{\left(x^4-5\right)^{\frac{1}{2}}}+\frac{3x^4-15}{(x^4-5)^{\frac{1}{2}}}[/tex][tex]y^{\prime}=\frac{6x^4+3x^4-15}{\left(x^4-5\right)^{\frac{1}{2}}}[/tex][tex]y^{\prime}=\frac{9x^4-15}{\left(x^4-5\right)^{\frac{1}{2}}}[/tex][tex]y^{\prime}=\frac{3(3x^4-5)}{\left(x^4-5\right)^{\frac{1}{2}}}[/tex][tex]y^{\prime}=\frac{3(3x^4-5)}{\sqrt{x^4-5}}[/tex][tex]y^{\prime}=\frac{3(3x^4-5)}{\sqrt{x^4-5}}\times\frac{\sqrt{x^4-5}}{\sqrt{x^4-5}}[/tex][tex]y^{\prime}=\frac{3(3x^4-5)\sqrt{x^4-5}}{x^4-5}[/tex]Final answer:
[tex]y^{\prime}=\frac{3(3x^4-5)\sqrt{x^4-5}}{x^4-5}[/tex]