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

Jan 26, 2018

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

At $\left(0 , 0\right)$, $\frac{\mathrm{dy}}{\mathrm{dx}} = 0$

Explanation:

When doing implicit differentiation, you follow these essential steps:

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

So, let's differentiate both sides:

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

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

$\frac{d}{\mathrm{dx}} \left[\tan \left(x + y\right)\right] \cdot \frac{d}{\mathrm{dx}} \left[x + y\right]$

This comes out to:

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

Putting this back in the whole equation:

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

Now, you just solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$ using some basic algebra:

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

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 - {\sec}^{2} \left(0\right)}{\sec} ^ 2 \left(0\right)$

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 - 1}{1} = 0$

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

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

Hope that helped :)

Jan 26, 2018

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

Explanation:

Implicit differentiation is just differentiation using chain rule.

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

Rearranging:

$1 + \frac{\mathrm{dy}}{\mathrm{dx}} = {\cos}^{2} \left(x + y\right)$
$\frac{\mathrm{dy}}{\mathrm{dx}} = {\cos}^{2} \left(x + y\right) - 1$

Substituting $\left(0 , 0\right)$ in above equation,
we get:
$\frac{\mathrm{dy}}{\mathrm{dx}} = {\cos}^{2} \left(0\right) - 1$
i.e $\frac{\mathrm{dy}}{\mathrm{dx}} = 0$