Yah
ok, so remember
x(dy/dx)=1
but
y(dy/dx)=dy/dx
solve for dy/dx as if it was a varaible
use chain rule and stuff
-33x¹⁰+2x²²y+y⁹=-30
take derivitive
-33(10)x⁹+2(22x²¹y+x²²dy/dx)+9y⁸dy/dx=0
distribute
-330x⁹+44x²¹y+2x²²dy/dx+9y⁸dy/dx=0
add 330x⁹ to both sides and minus 44x²¹y from both sides
2x²²dy/dx+9y⁸dy/dx=330x⁹-44x²¹y
undistribute dy/dx
dy/dx(2x²²+9y⁸)=330x⁹-44x²¹y
divide both sides by (2x²²+9y⁸)
your calculation is incorrect, see it should be 2x^22, not x^22
anyway
taht is the slope at taht point
remember point slope
y-y1=m(x-x1)
we needs to find slope
at point (1,1)
x=1, y=1
dy/dx=26
at point (1,1)
y-1=26(x-1)
y-1=26x-26
y=26x-25 is de equation
ok, so remember
x(dy/dx)=1
but
y(dy/dx)=dy/dx
solve for dy/dx as if it was a varaible
use chain rule and stuff
-33x¹⁰+2x²²y+y⁹=-30
take derivitive
-33(10)x⁹+2(22x²¹y+x²²dy/dx)+9y⁸dy/dx=0
distribute
-330x⁹+44x²¹y+2x²²dy/dx+9y⁸dy/dx=0
add 330x⁹ to both sides and minus 44x²¹y from both sides
2x²²dy/dx+9y⁸dy/dx=330x⁹-44x²¹y
undistribute dy/dx
dy/dx(2x²²+9y⁸)=330x⁹-44x²¹y
divide both sides by (2x²²+9y⁸)
your calculation is incorrect, see it should be 2x^22, not x^22
anyway
taht is the slope at taht point
remember point slope
y-y1=m(x-x1)
we needs to find slope
at point (1,1)
x=1, y=1
dy/dx=26
at point (1,1)
y-1=26(x-1)
y-1=26x-26
y=26x-25 is de equation