To find the derivative through implicit differentiation, we have to take the derivative of every term with respect to x.  Don't forget that each time you take the derivative of a term containing y, you must multiply its derivative by y'.  So, when we take the derivative of each term, we get   The next step is to solve for y', so we put all terms containing y' on the left side of the equation: .  To get y' alone, divide both sides by  to get .  To simplify even further, we can factor a 2 out of the numerator and denominator and cancel them.  So, the final answer is .

To find the derivative through implicit differentiation, we have to take the derivative of every term with respect to x.  Don't forget that each time you take the derivative of a term containing y, you must multiply its derivative by y'.  So, when we take the derivative of each term, we get .  The next step is to solve for y', so we put all terms containing y' on the left side of the equation: .  To get y' alone, divide both sides by -3 to get .  To simplify even further, we can factor a -2 out of the numerator and denominator and cancel them.  So, the final answer is .

To find the derivative through implicit differentiation, we have to take the derivative of every term with respect to x.  Don't forget that each time you take the derivative of a term containing y, you must multiply its derivative by y'.  So, when we take the derivative of each term, we get .  The next step is to solve for y', so we put all terms containing y' on the left side of the equation: .  Next, factor out the y' from both terms on the left side of the equation so that we can solve for it: .  To get y' alone, divide both sides by  to get a final answer of  .  

Before we take the derivative of the logarithmic function, we can make it easier for ourselves by simplifying the equation to .  We can bring the exponent of 6 down in front of the natural log of x due to properties of logarithms.  Next, take the derivative of each term in terms of x.  Don't forget to multiply by y' each time you take the derivative of a term containing y!  When we do this, we should get  because the derivative of lnx is 1/x.  Next, solve for y' by multiplying both sides by y to get the final answer of .

To take the derivative, you must first take the derivative of the outside function, which is sine.  However, the , or the angle of the function, remains the same until we take its derivative later.  The derivative of sinx is cosx, so you the first part of  will be .  Next, take the derivative of the inside function, .  Its derivative is , so by the chain rule, we multiply the derivatives of the inside and outside functions together to get .

By the chain rule, we must first take the derivative of the outside function by bringing the power down front and reducing the power by one.  When we do this, we do not change the function that is in the parentheses, or the inside function.  That means that the first part of  will be .  Next, we must take the derivative of the inside function.  Its derivative is .  The chain rule says we must multiply the derivative of the outside function by the derivative of the inside function, so the final answer is .

To differentiate the equation above, start by applying the derivative operation to both sides, 

Both sides will require the product rule to differentiate, 

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 Common Mistake 

A common mistake in the previous step would be to conclude that instead of  . The former is not correct; if we were looking for the derivative with respect to , then  would in fact be . But we are not differentiating with respect to , we're looking for the derivative with respect to . 

We are assuming that  is a function of , so we must apply the chain rule by differentiating with respect to  and multiplying by the derivative of  with respect to  to obtain . 

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Collect terms with a derivative onto one side of the equation, factor out the derivative, and divide out  to solve for the derivative , 

Therefore, 

                          (1)

An easier way to think about this:

Because  is a function of a function, we must apply the chain rule. This can be confusing at times especially for function like equation (1). The differentiation is easier to follow if you use a substitution for the inner function, 

Let,

                               (2)

So now equation (1) is simply, 

                               (3)

Note that  is a function of . We must apply the chain rule to find  , 

                            (4)

To find the derivatives on the right side of equation (4), differentiate equation (3) with respect to , then Differentiate equation (2) with respect to . 

Substitute into equation (4),  

                  (5)

Now use  to write equation (5) in terms of  alone: 

Here we use the chain rule: 

Let 

Then 

And 

Using the chain rule, we have

.

Hence, .

Notice that we could have also simplified  first by cancelling the natural log and the exponential function leaving us with just , thereby avoiding the chain rule altogether.