Derivatives and slope when x =1

The right-hand end of the lines of text have not been included in your image. I *think* the following is what is meant:

Find the slope, at any point, of the curve whose equation is [imath]\frac{x^2}{3^2} + \frac{y^2}{2^2} = 1[/imath]; and give the numerical value of the slope at any place where [imath]x = 0[/imath], and at that where [imath]x = 1[/imath].
Click to expand...

Is the above correct? When you reply, please include a clear listing of your thoughts and efforts so far, beginning with your derivative of the equation (which involves nothing more than polynomial terms). Thank you.

Eliz.
 
Alternatively, do you even need calculus to find the slope of a tangent line to a circle???
 
khansaheb said:
x^2 + y^2 = a

Where 'a' is a constant and ^ means "square".
Click to expand...
There is a convenient formatting option available on the forum when composing your posts (if you don't wish to learn LaTex).

If you go all the way to the right hand side of the icons above and click on the three vertical dots there, a new series of icons appears underneath, one of which is "X1" and another "X1".

Using the "X1" will save you having to explain what "^" means (it won't always mean "square", of course) and your

x^2 + y^2 = a​
easily becomes
x2 + y2 = a​

When wishing to show a term raised to some power just highlight the index and click on "X1". ?
 
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