Francis Oliver

2022-10-11

Find $\frac{dy}{dx}$ by implicit differentiation.

$\mathrm{tan}(x+y)=x$

So far, I got to

${\mathrm{sec}}^{2}(x+y)(1+\frac{dy}{dx})=1$

but then im lost.. can someone please help and explain? I would really appreciate it!!

$\mathrm{tan}(x+y)=x$

So far, I got to

${\mathrm{sec}}^{2}(x+y)(1+\frac{dy}{dx})=1$

but then im lost.. can someone please help and explain? I would really appreciate it!!

dkmc4175fl

Beginner2022-10-12Added 15 answers

Well, you have ${\mathrm{sec}}^{2}(x+y)(1+{y}^{\prime})=1$, from which you get ${y}^{\prime}={\mathrm{cos}}^{2}(x+y)-1$.

Elise Kelley

Beginner2022-10-13Added 2 answers

you need to foil sec${}^{2}(x+y)(1+{y}^{\prime}(x))=1$ and then solve for ${y}^{\prime}(x)$.

Think about sec${}^{2}(x+y)$ as one whole.

Hint: the answer will be in terms of $x$ and $y$

Think about sec${}^{2}(x+y)$ as one whole.

Hint: the answer will be in terms of $x$ and $y$

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