Question

x^2 + 4y^5 = 53 calculate the derivative at the point (7,1)
I'm assuming you reorder the equation to y = blah. but I'm not sure, please show/explain

Answer 1

you could do that since its an odd exponent on the y (the fifth root would still leave you with one function), or you could also do implicit differentiation and then solve for dy/dx and plugging in y again.

Answer 2

actually i guess you could leave y in with the equation as dy/dx = and just plug in the y value as well as the x value

Answer 3

so 4y^5/53-x^2... and then use the quotient rule?

Answer 4

then plug in for x and y?

Answer 5

\[ \begin{split}
x^{2} + 4y^{5} &= 53\\
\frac{d}{dx}(x^{2} + 4y^{5}) &= \frac{d}{dx}(53)\\
\frac{d}{dx}(x^{2}) + \frac{d}{dx}(4y^{5}) &= 0\\
2x + 4*5y^{4}\frac{dy}{dx} &= 0\\
2x + 20y^{4}\frac{dy}{dx} &= 0\\
&x=1, y=7\\
2(7) + 20(1)^{4}\frac{dy}{dx} &= 0\\
14 + 20\frac{dy}{dx} &= 0\\
20\frac{dy}{dx} = -14\\
\frac{dy}{dx} &= -7/10\\
\end{split} \]

Answer 6

oops, should have said "x=7, y=1"

Answer 7

That's it, so basically take the derivative of both of them, and get rid of all of the ones without x's or y's and then plug in, got it