Answer :
Final answer:
The derivative of y with respect to x, y', is given by the equation y' = y + xy' + 2y * (dy/dx) - 1/sqrt(1 - (sinx4)^2) * (d(sinx4)/dx) + tan-1(x3-1) * (dx/dx) + x * (d(tan-1(x3-1))/dx) + 1/(x2+x) * (d(x2+x)/dx) + (dy/dx) - 12x^3
Explanation:
To find y', we need to differentiate each term of the given equation separately and then combine the results.
Let's differentiate each term:
- yx: Using the product rule, the derivative of yx with respect to x is y * (dx/dx) + x * (dy/dx) = y + xy'
- y2: Using the power rule, the derivative of y2 with respect to x is 2y * (dy/dx)
- cos-1(sinx4): Using the chain rule, the derivative of cos-1(sinx4) with respect to x is -1/sqrt(1 - (sinx4)^2) * (d(sinx4)/dx)
- xtan-1(x3-1): Using the product rule and chain rule, the derivative of xtan-1(x3-1) with respect to x is tan-1(x3-1) * (dx/dx) + x * (d(tan-1(x3-1))/dx)
- log5(x2+x): Using the chain rule, the derivative of log5(x2+x) with respect to x is 1/(x2+x) * (d(x2+x)/dx)
- y: The derivative of y with respect to x is (dy/dx)
- -3x^4: Using the power rule, the derivative of -3x^4 with respect to x is -12x^3
Now, let's combine the results:
y' = y + xy' + 2y * (dy/dx) - 1/sqrt(1 - (sinx4)^2) * (d(sinx4)/dx) + tan-1(x3-1) * (dx/dx) + x * (d(tan-1(x3-1))/dx) + 1/(x2+x) * (d(x2+x)/dx) + (dy/dx) - 12x^3
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