Science:Math Exam Resources/Courses/MATH105/April 2018/Question 01 (d)

MATH105 April 2018

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Question 01 (d)
(d) Let ${\textstyle f(x,y)=y^{3}\sin(2x)}$. Find the second partial derivative ${\textstyle f_{xy}}$.

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 hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Recall that ${\textstyle f_{xy}={\frac {\partial ^{2}f}{\partial y\partial x}}={\frac {\partial }{\partial y}}\left({\frac {\partial f}{\partial x}}\right).}$

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. First we calculate the partial derivative ${\textstyle f_{x}}$: $$f_{x}={\frac {\partial }{\partial x}}{\bigg (}y^{3}\sin(2x){\bigg )}=y^{3}{\frac {\partial }{\partial x}}(\sin(2x))=2y^{3}\cos(2x).$$ Note that when we take the partial derivative ${\frac {\partial }{\partial x}}$, we consider $y$ as a constant value. Then we get $$f_{xy}={\frac {\partial }{\partial y}}{\bigg (}2y^{3}\cos(2x){\bigg )}={\frac {\partial }{\partial y}}(2y^{3})\cos(2x)=6y^{2}\cos(2x).$$ Answer: $\color {blue}f_{xy}=6y^{2}\cos(2x).$