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Sagot :
Solution:
Given the equation:
[tex]y=6\sqrt{x+14}\text{ ----- equation 1}[/tex]Equation 1 can be expressed in a simpler manner to be
[tex]y=6(x+14)^{\frac{1}{2\text{ }}}----\text{ equation 2}[/tex]To find
[tex]\frac{dy}{dx}[/tex]we take the derivative of y with respect to x.
From the product rule,
[tex]\frac{d(UV)}{dx}=U\frac{dV}{dx}+V\frac{dU}{dx}[/tex]In this case,
[tex]\begin{gathered} U=6 \\ V=(x+14)^{\frac{1}{2}} \end{gathered}[/tex]thus,
[tex]\frac{d(6(x+14)^{\frac{1}{2}})}{dx}=6\frac{d((x+14)^{\frac{1}{2}})}{dx}+(x+14)^{\frac{1}{2}}\frac{d(6)}{dx}[/tex][tex]\begin{gathered} \frac{d((x+14)^{\frac{1}{2}})}{dx}=\frac{1}{2}(x+14)^{-\frac{1}{2}}\times1 \\ =\frac{1}{2}(x+14)^{-\frac{1}{2}} \\ \frac{d(6)}{dx}=0 \\ The\text{ derivative of a constant gives zero} \end{gathered}[/tex]Substituting these parameters into the product rule equation, we have
[tex]\begin{gathered} \frac{d(6(x+14)^{\frac{1}{2}})}{dx}=6(\frac{1}{2}(x+14)^{-\frac{1}{2}})+(x+14)^{\frac{1}{2}}(0) \\ =3(x+14)^{-\frac{1}{2}}+0 \\ thus, \\ \frac{d(6\sqrt{x+14})}{dx}=\frac{3}{\sqrt{x+14}} \end{gathered}[/tex]Hence, the derivative is expressed as
[tex]\frac{3}{\sqrt{x+14}}[/tex]
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