using implicit differentiation find y' of x^2 -3^y +7xy=8y^5. I understand everything except for the 3^y part. PLEASE HELP!

Answer 1

Here e is the base of the natural logarithm rounded to 2.718281828.

From x² − 3^y +7xy = 8y⁵, obtain:

x² − (e^{ln 3})^y + 7xy = 8y⁵.

Then implicit differentiation gives

2x(dx/dx) − [ln 3 • e^{yln 3} • dy/dx] + d(7x)/dx • y + 7x • dy/dx = 5 • 8y⁴(dy/dx).

Rewrite as 2x − ln 3(e^{yln 3}) y' + 7y + 7xy' = 40y⁴y'.

Express as 2x + 7y = 40y⁴y' + ln 3(e^{yln 3}) y' − 7xy'.

Finally isolate y' as [2x + 7y] ÷ [40y⁴ + ln 3(e^{yln 3}) − 7x].