Question

use implicit differentiation to find the derivative of:

Answer 1

\[x^2y^4+\tan(y)=8x+1\]

Answer 2

\[\frac{ d }{ dx }(x^2y^4 + \tan(y) = 8x + 1)\]At first using the product rule on x^2y^4,\[(2xy^4 + x^2(\frac{ d }{ dx }(y^4))) + \frac{ d }{ dx }\tan(y) = 8\]\[2xy^4 + x^2(4y^3y') + \frac{ d }{ dx }\tan(y) = 8\]

Answer 3

To differentiate tan(y), the chain rule will need to be used.\[h(x) = f(g(x))\]\[h'(x) = f'(g(x))g'(x)\]Where f(x) = tan(x), and g(x) = y\[h(x) = \tan(y(x))\]\[h'(x) = \sec^2(y(x))\frac{ d }{ dx }(y(x))\]

Answer 4

If you're confused about where the h(x) came from, I was just showing that the chain rule is needed for differentiating that part.

Answer 5

Back to the problem at hand.\[2xy^4 + 4y^3y'x^2 + \sec^2(y)y' = 8\]\[y'(4y^3x^2 + \sec^2(y)) = 8 - 2xy^4\]\[y' = \frac{ 8 - 2xy^4 }{ 4y^3x^2 + \sec^2(y) }\]

Answer 6

makes sense!! thank you!!