# Implicit differentiation.

• Nov 3rd 2013, 11:17 PM
sepoto
Implicit differentiation.
I am being asked to find $\frac{dy}{dx}$ by implicit differentiation of the function $y=x^{1/n}$. I'm really not sure where to begin yet with this problem. I watched a few videos on implicit differentiation but I'm still new to the process. I'm hoping to get some pointers on how to start working this problem.

• Nov 4th 2013, 01:05 AM
chiro
Re: Implicit differentiation.
Hey sepoto.

In this case you don't need to use implicit differentiation since you have y = f(x).

But for your information implicit differentiation usually involves differentiating both sides and then putting dy/dx on one side and the rest on the other.
• Nov 4th 2013, 02:22 AM
ibdutt
Re: Implicit differentiation.
Firstly we should know what are implicit functions. in simple words implicit function is a function where in we cannot express one variable it terms of another e.g., xy + x^2y + 3y + 4x = 0; etc.
In such cases differentiate the function, we get
y + x dy/dx + 2xy + x^2 dy/dx + 3 dy/dx + 4 = 0
Now collect all terms containing dy/dx on one side and find the value of dy/dx;
in the example under consideration we will get
( x + x^2 + 3 ) dy/dx = -y-4
dy/dx = -(y+4)/( x + x^2 + 3 )
• Nov 5th 2013, 02:15 PM
SlipEternal
Re: Implicit differentiation.
Raise both sides to the $n$-th power. So, $y^n = (y)^n = (x^{1/n})^n = x^{n/n} = x$. Then, take the derivative of both sides:

$\dfrac{d}{dx}(y^n) = \dfrac{d}{dx}(x)$

On the LHS, you need to apply the chain rule since you are differentiating with respect to $x$. On the RHS, you just get 1.

$\dfrac{d}{dy}(y^n) \dfrac{dy}{dx} = 1$

Now, you can differentiate with respect to $y$ and get:

$ny^{n-1} \dfrac{dy}{dx} = 1$

Solving for $\dfrac{dy}{dx}$, you have:

$\dfrac{dy}{dx} = \dfrac{1}{ny^{n-1}}$

From your initial equation, you have $y = x^{1/n}$. Plugging that in, you have:

$\dfrac{dy}{dx} = \dfrac{1}{n(x^{1/n})^{n-1}} = \dfrac{1}{nx^{(n-1)/n}} = \dfrac{1}{n} x^{-(n-1)/n} = \dfrac{1}{n} x^{(1/n-1)}$

This shows that the Power Rule applies even when the exponent of $x$ is a rational number, not just an integer. You won't see a proof that the power rule works when the exponent for $x$ is any real number until you get to logarithmic and exponential differentiation.