Maths A-level dy/dx =.... dy/dx = 0?
I need some help here. I am stuck on that devide all by x^2
Anyway here is a question dy/dx = 8^3 +1 / x^2 find value for dy/dx = 0
Could you please go throught the workings out especially what you do what that / by x^2
Update:and from the mark scheme the answer is -0.5
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Answers & Comments
herp derp 8x^3 changes is quite a bit.
0=8x^3+1/x^2
multiply both sides by x^2
0=8x^5 + 1
x^5=-1/8
x=(-1/8)^-5
Wow, you need to disambiguate your question.
You should have written the quantity 8x^3 +1 in parentheses: (8x^3+1)/x^2
Otherwise, you must assume that 8x^3 is a separate term from the 1/x^2, which gave me the wrong answer.
If you're going to ask for help on your homework, make sure you understand the basics of expressing statements unambiguously.
8x^3 + 1 / x^2 = 0
Multiply by x^2 to both sides
8x^3 + 1 = 0
Minus 1 to both sides
8x^3 = -1
Divide both sides by 8
x^3 = -1/8
Cube root both sides
x= (-1/8)^1/3
= -0.5
r u sure its 8^3 and not 8x^3