How do you find the derivative of #tan(x/y)=x+y#?
2 Answers
Explanation:
#"differentiate "tan(x/y)" using the "color(blue)"chain rule"#
#rArrd/dx(tan(x/y))#
#=sec^2(x/y)xxd/dx(x/y)#
#"differentiate "x/y" using the "color(blue)"quotient rule"#
#=sec^2(x/y)xx(y-x.dy/dx)/y^2#
#=(ysec^2(x/y)-xsec^2(x/y)dy/dx)/y^2#
#"returning to the original"#
#(ysec^2(x/y)-xsec^2(x/y)dy/dx)/y^2=1+dy/dx#
#rArrysec^2(x/y)-xsec^2(x/y)dy/dx=y^2+y^2dy/dx#
#rArrdy/dx(-xsec^2(x/y)-y^2)=y^2-ysec^2(x/y)#
#rArrdy/dx=-(y^2-ysec^2(x/y))/(xsec^2(x/y)+y^2)#
Explanation:
You need the chain rule on the tangent part:
Distribute on the left side:
Move everything with a
Factor out the
Get
Get a common denominator in both the numerator and the denominator:
Simplify the complex fraction (I think of it as keep, change, turn):
I factored the negative out of the denominator and distributed it to the numerator:
I decided to factor the common