Square Root Diff Example

[Jason76]

Post Edited

\(\displaystyle y = \sqrt{7 + 2e^{9x}}\)

\(\displaystyle y = (7 + 2e^{9x})^{1/2}\)

\(\displaystyle y = \dfrac{1}{2}(u)^{-1/2} du\)

\(\displaystyle y = \dfrac{1}{2}(u)^{-1/2} (2e^{v} dv)\)

\(\displaystyle y = \dfrac{1}{2}(u)^{-1/2} (2e^{v} (9)\)

\(\displaystyle y = \dfrac{1}{2}(u)^{-1/2} (18e^{v}\)

\(\displaystyle y = \dfrac{1}{2}(u)^{-1/2} (18e^{9x}\)

\(\displaystyle y = \dfrac{1}{2}(7 + 2e^{9x})^{-1/2} (18e^{9x}\)

 

[Deleted member 4993]

\(\displaystyle y = \sqrt{7 + 2e^{9x}}\)

\(\displaystyle y = (7 + 2e^{9x})^{1/2}\)

\(\displaystyle y = \dfrac{1}{2}(u)^{-1/2} du\)

\(\displaystyle y = \dfrac{1}{2}(u)^{-1/2} (2e^{9x})\)............ Incorrect .......\(\displaystyle \frac{d}{dx}[7+2e^{9x}] \ \ne \ 2e^{9x}\)

\(\displaystyle y = \dfrac{1}{2}(7 + 2e^{9x})^{-1/2} (2e^{9x})\)

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[Jason76]

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What is \(\displaystyle y' = \dfrac{d}{dx} (7 + 2e^{9x})\) ? Hint - What should I first look at?

 

[Jason76]

\(\displaystyle y = \sqrt{7 + 2e^{9x}}\)

\(\displaystyle y = (7 + 2e^{9x})^{1/2}\)

\(\displaystyle y = \dfrac{1}{2}(u)^{-1/2} du\)

\(\displaystyle y = \dfrac{1}{2}(u)^{-1/2} [2e^{v} dv]\)

\(\displaystyle y = \dfrac{1}{2}(u)^{-1/2} [2e^{v} (9)]\)

\(\displaystyle y = \dfrac{1}{2}(u)^{-1/2} [18e^{v}]\)

\(\displaystyle y = \dfrac{1}{2}(u)^{-1/2} [18e^{9x}]\)

\(\displaystyle y = \dfrac{1}{2}(7 + 2e^{9x})^{-1/2} [18e^{9x}]\)