Answer:
[tex]\frac{3}{2} (3x+16)^{\frac{-1}{2} }[/tex]
Explanation:
Using Chain Rule:
[tex]\sqrt{3x+16} =(3x+16)^\frac{1}{2} \\\frac{d[(3x+16)^\frac{1}{2}]}{dx} =\frac{1}{2} (3x+16)^\frac{-1}{2} 3\\=\frac{3}{2} (3x+16)^\frac{-1}{2}[/tex]
[tex]\boxed{\bf\: \dfrac{d}{dx} \sqrt{3x + 16} = \dfrac{3}{2 \sqrt{3x + 16} } \: } \\ [/tex]
Consider,
[tex]\sf\: \dfrac{d}{dx} \sqrt{3x + 16} \\ [/tex]
[tex]\sf\: = \: \dfrac{d}{dx} {\left[ 3x + 16\right]}^{ \frac{1}{2} } \\ [/tex]
We know,
[tex]\sf\:\boxed{\begin{aligned}& \qquad \:\sf \: \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1} \qquad \: \\ \\& \qquad \:\sf \: \dfrac{d}{dx}f\left[ g(x)\right] = f'\left[ g(x)\right]\dfrac{d}{dx}g(x) \: \qquad \end{aligned}} \\ [/tex]
So, Using these results, we get
[tex]\sf\: = \: \dfrac{1}{2} {\left[3x + 16 \right]}^{ \frac{1}{2} - 1} \: \dfrac{d}{dx} (3x + 16) \\ [/tex]
[tex]\sf\: = \: \dfrac{1}{2} {\left[3x + 16 \right]}^{ - \frac{1}{2}} \:\left[ 3 \dfrac{d}{dx}x + \dfrac{d}{dx} 16\right]\\ [/tex]
[tex]\sf\: = \: \dfrac{1}{2 \sqrt{3x + 16} } \:\left[ 3 \times 1 + 0\right]\\ [/tex]
[tex]\sf\: = \: \dfrac{1}{2 \sqrt{3x + 16} } \:\left[ 3\right]\\ [/tex]
[tex]\sf\: = \: \dfrac{3}{2 \sqrt{3x + 16} }\\ [/tex]
Hence,
[tex]\implies\sf\:\boxed{\bf\: \dfrac{d}{dx} \sqrt{3x + 16} = \dfrac{3}{2 \sqrt{3x + 16} } \: } \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Additional Information:
[tex]\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {x}^{n} & \sf {nx}^{n - 1}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}[/tex]
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Answers & Comments
Answer:
[tex]\frac{3}{2} (3x+16)^{\frac{-1}{2} }[/tex]
Explanation:
Using Chain Rule:
[tex]\sqrt{3x+16} =(3x+16)^\frac{1}{2} \\\frac{d[(3x+16)^\frac{1}{2}]}{dx} =\frac{1}{2} (3x+16)^\frac{-1}{2} 3\\=\frac{3}{2} (3x+16)^\frac{-1}{2}[/tex]
Answer:
[tex]\boxed{\bf\: \dfrac{d}{dx} \sqrt{3x + 16} = \dfrac{3}{2 \sqrt{3x + 16} } \: } \\ [/tex]
Explanation:
Consider,
[tex]\sf\: \dfrac{d}{dx} \sqrt{3x + 16} \\ [/tex]
[tex]\sf\: = \: \dfrac{d}{dx} {\left[ 3x + 16\right]}^{ \frac{1}{2} } \\ [/tex]
We know,
[tex]\sf\:\boxed{\begin{aligned}& \qquad \:\sf \: \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1} \qquad \: \\ \\& \qquad \:\sf \: \dfrac{d}{dx}f\left[ g(x)\right] = f'\left[ g(x)\right]\dfrac{d}{dx}g(x) \: \qquad \end{aligned}} \\ [/tex]
So, Using these results, we get
[tex]\sf\: = \: \dfrac{1}{2} {\left[3x + 16 \right]}^{ \frac{1}{2} - 1} \: \dfrac{d}{dx} (3x + 16) \\ [/tex]
[tex]\sf\: = \: \dfrac{1}{2} {\left[3x + 16 \right]}^{ - \frac{1}{2}} \:\left[ 3 \dfrac{d}{dx}x + \dfrac{d}{dx} 16\right]\\ [/tex]
[tex]\sf\: = \: \dfrac{1}{2 \sqrt{3x + 16} } \:\left[ 3 \times 1 + 0\right]\\ [/tex]
[tex]\sf\: = \: \dfrac{1}{2 \sqrt{3x + 16} } \:\left[ 3\right]\\ [/tex]
[tex]\sf\: = \: \dfrac{3}{2 \sqrt{3x + 16} }\\ [/tex]
Hence,
[tex]\implies\sf\:\boxed{\bf\: \dfrac{d}{dx} \sqrt{3x + 16} = \dfrac{3}{2 \sqrt{3x + 16} } \: } \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Additional Information:
[tex]\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {x}^{n} & \sf {nx}^{n - 1}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}[/tex]