Find the slope of the tangent to the curve x^4+xy+y^2=57 at (1,7) ?

Question

ikram002p · Answer

differentiat for x.
then solve for dx\dy
then find dx\dy(7)  by satisfy x=7, y=1
wat steps of these u r stack in ??

anonymous · Answer

In everything, I already found the derivative and solved for dy/dx and got the wrong answer several times. I've been working on this for a very long time, I really just need it explained step by step.

ikram002p · Answer

show me ur work

anonymous · Answer

Found the derivative 4x^3+x+xy+2y = 0. Solved for dy/dx and got dy/dx = -4x^3-y/x+2y and then plugged in (1,7) and got -71/15 the first time and 34/15 another time and -65/21 a third time.

anonymous · Answer

I also tried 

4 x^3 + y + x dy /dx + 2 y dy /dx = 0 . 
dy / dx [ x + 2 y ] = - [ y + 4 x^3 ] 
dy / dx = - [ y + 4 x^3 ] / [ x + 2 y ] 
At ( 1 , 7 ) dy / dx = - [ 7 + 4 ] / [ 1 + 14 ] = - 11 / 15.

anonymous · Answer

And 

4 x^3 + y + x dy/dx + 2y dy/dx = 0, so dy/dx ( x + 2 y) = - 4 x^3 

dy/dx = - 4 x^3 / (x + 2 y) 

dy/dx = -4 /(1 + 14) = - 4 /15

anonymous · Answer

Is either -11/15 or -4/15 right?

ikram002p · Answer

u've got the sol  !! its -11\15
and ur method is correct , but the second sol is wrong :)

anonymous · Answer

Thank you very much for your help!