Question

y^6+x^3=y^2+10x ,normal at(0,1) at the given point,find the slope of the curve,the line that is tangent to the curve,or the line that is normal to the curve,as requested.

Answer 1

can you be more clear about the question?

Answer 2

at the given point,find the slope of the curve,the line that is tangent to the curve,or the line that is normal to the curve,as requested.

Answer 3

\[y ^{6}+x ^{3}=y ^{2}+10x , normal at (0,1)\]

Answer 4

I would write the equation as
\[ y^6+x^3-y^2-10x=0 \]
and use implicit differentiation. solve for dy/dx . can you do that ?

Answer 5

\[6y ^{5}+3x ^{2}-2y-10=0\]

Answer 6

you left out an important part.

when you take the derivative of x with respect to x you get
\[ \frac{d }{dx}x = \frac{dx}{dx}= 1 \]

when you take the derivative of y with respect to x you get
\[ \frac{d }{dx}y = \frac{dy}{dx} \]

if you have a function of y , for example \(y^2\), you must use the chain rule:
\[ \frac{d }{dx}y^2 =2 y \frac{d }{dx}y =2 y \frac{dy }{dx}\]

Answer 7

people often abbreviate \( \frac{dy}{dx} \) as y' (easier to type)
The derivative is
\[ 6y ^{5} y' +3x ^{2}-2y\ y'-10=0 \]

Answer 8

now solve for y'
(you could solve for y' in terms of x and y , or you could first sub in x=0, y=1 and then solve)