Science:Math Exam Resources/Courses/MATH110/December 2015/Question 06 (a)

MATH110 December 2015

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Question 06 (a)
Find $y'$ if $y={\sqrt {x+3x^{2}}}.$

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
Consider the implicit differentiation.

Hint 2
(Alternative way) Use chain rule.

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Solution 1

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Since $y={\sqrt {x+3x^{2}}}$ implies that $y^{2}=x+3x^{2}$, we apply the implicit differentiation to the second equation. Then, we get ${\frac {d}{dx}}(y^{2})={\frac {d}{dx}}(x+3x^{2})$ Using the chain rule, the left hand side of the equation becomes $2yy'$. On the other hand, by the power rule, the right hand side can be simplified easily as ${\frac {d}{dx}}(x+3x^{2})=1+6x$. Collecting the terms, we obtain $2yy'=1+6x$. Then, plugging $y={\sqrt {x+3x^{2}}}$ back, we have $y'={\frac {1+6x}{2y}}=\color {blue}{\frac {1+6x}{2{\sqrt {x+3x^{2}}}}}$.

Solution 2

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. The below solution is more direct. We put $u=x+3x^{2}$ and use chain rule to get the derivative $y'$; $y'={\frac {dy}{dx}}={\frac {dy}{du}}{\frac {du}{dx}}={\frac {d}{du}}{\sqrt {u}}{\frac {d}{dx}}(x+3x^{2})={\frac {1}{2{\sqrt {u}}}}(1+6x)={\frac {1+6x}{2{\sqrt {x+3x^{2}}}}}$. Therefore, we obtain the derivative $\color {blue}y'={\frac {1+6x}{2{\sqrt {x+3x^{2}}}}}$.