Question 237326
You need to implicit differentiation for this problem and solve for {{{dy/dx}}}
So we have.
{{{x^2+y^2=25}}}
The derivative is {{{2x+2y*dy/dx=0}}} the 2y produces the {{{dy/dx}}} because y is a function so you have to apply the chain rule to it.
So now.
{{{2y*dy/dx=-2x}}}
{{{dy/dx=-2x/2y}}}
{{{dy/dx=-x/y}}}
And x and y are given by (3,4) giving x=3 and y=4. ({{{x[1]}}},{{{y[1]}}})
So {{{dy/dx=-(3/4)}}}
Now for the tangent line you have.
{{{y-y[1]=(dy/dx)(x-x[1])}}}
so we have {{{y-4=-(3/4)(x-3)}}}
Giving us {{{y-4=-(3/4)x+9/4}}} Add 4 to both sides. 4 can be rewritten as {{{16/4}}}
{{{y=-(3/4)x+9/4+16/4}}}
{{{y=-(3/4)x+25/4}}}
And that is your tangent line.