How do you find $\frac{d y}{d x}$ $dy/dx$ by implicit differentiation of $tan (x + y) = x$ $tan(x+y)=x$ and evaluate at point (0,0)?

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How do you find $\frac{d y}{d x}$ $dy/dx$ by implicit differentiation of $tan (x + y) = x$ $tan(x+y)=x$ and evaluate at point (0,0)?

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Answer 1 · 2018-01-26T16:03:52

$\frac{d y}{d x} = \frac{1 - {sec}^{2} (x + y)}{{sec}^{2} (x + y)}$ $dy/dx = [1-sec^2(x + y)]/sec^2(x + y)$

At $(0, 0)$ $(0,0)$ , $\frac{d y}{d x} = 0$ $dy/dx = 0$

Explanation:

When doing implicit differentiation, you follow these essential steps:

Take the derivative of both sides of the equation with respect to $x$ $x$ .
Differentiate terms with $x$ $x$ as normal.
Differentiate terms with $y$ $y$ as normal too but tag on a $\frac{d y}{d x}$ $dy/dx$ to the end.
Solve for the $\frac{d y}{d x}$ $dy/dx$ .

So, let's differentiate both sides:

$\frac{d}{d x} [tan (x + y)] = \frac{d}{d x} [x]$ $d/dx[tan(x + y)] = d/dx[x]$

The right hand side just comes out as $1$ $1$ , but the left hand side will require that we use a chain rule. This goes as:

$\frac{d}{d x} [tan (x + y)] \cdot \frac{d}{d x} [x + y]$ $d/dx[tan(x + y)] * d/dx[x + y]$

This comes out to:

${sec}^{2} (x + y) \cdot (1 + \frac{d y}{d x})$ $sec^2(x + y) * (1 + dy/dx)$

Putting this back in the whole equation:

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

Now, you just solve for $\frac{d y}{d x}$ $dy/dx$ using some basic algebra:

$\frac{d y}{d x} = \frac{1 - {sec}^{2} (x + y)}{{sec}^{2} (x + y)}$ $dy/dx = [1-sec^2(x + y)]/sec^2(x + y)$

Now, you're given the point $(0, 0)$ $(0,0)$ to evaluate the derivative at. All you do is plug this in:

$\frac{d y}{d x} = \frac{1 - {sec}^{2} (0)}{{sec}^{2} (0)}$ $dy/dx = [1-sec^2(0)]/sec^2(0)$

$sec (0)$ $sec(0)$ is simply $\frac{1}{cos (0)}$ $1/cos(0)$ . Since $cos (0)$ $cos(0)$ = 1, $sec (0)$ $sec(0)$ is also 1. So, wherever we see $sec (0)$ $sec(0)$ , we just plug in $1$ $1$ :

$\frac{d y}{d x} = \frac{1 - 1}{1} = 0$ $dy/dx = [1-1]/1 = 0$

So your tangent line would have a slope of $0$ $0$ at the point $(0, 0)$ $(0,0)$ .

If you want more help in implicit differentiation, check out my video:

Hope that helped :)

Answer 2 · 2018-01-26T16:07:02

$\frac{d y}{d x} = {cos}^{2} (x + y) - 1$ $dy/dx = cos^2(x+y)-1$
Evaluating at $(0, 0)$ $(0,0)$ gives $\frac{d y}{d x} = 0$ $dy/dx=0$

Explanation:

Implicit differentiation is just differentiation using chain rule.

$\frac{d tan (x + y)}{d (x + y)} \times \frac{d (x + y)}{d x}$ $(d tan(x+y))/(d(x+y)) xx (d (x+y))/dx$ = $\frac{d x}{d x}$ $dx/dx$
${sec}^{2} (x + y) \times (1 + \frac{d y}{d x}) = 1$ $sec^2(x+y)xx(1+dy/dx) = 1$

Rearranging:

$1 + \frac{d y}{d x} = {cos}^{2} (x + y)$ $1+dy/dx = cos^2(x+y)$
$\frac{d y}{d x} = {cos}^{2} (x + y) - 1$ $dy/dx = cos^2(x+y)-1$

Substituting $(0, 0)$ $(0,0)$ in above equation,
we get:
$\frac{d y}{d x} = {cos}^{2} (0) - 1$ $dy/dx = cos^2(0)-1$
i.e $\frac{d y}{d x} = 0$ $dy/dx=0$

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How do you find $\frac{d y}{d x}$ $dy/dx$ by implicit differentiation of $tan (x + y) = x$ $tan(x+y)=x$ and evaluate at point (0,0)?

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