How to Calculate Implicit Derivatives: A Guide for Understanding

[Runaway]

In short, I need to know how do you do them.
I missed class, and our textbook is so bad that it might as well be written in a foreign language. I understand how to do dy/dx of an equation not in the form y =... ex. y^2 = x^3 + 2x + 5, (y'=(3x^2 +2)/2y) for example, but how would you take the derivative with respect to a variable not in the equation, t for example, and how would you take dx/dy for an equation?

 

[HallsofIvy]

In short, I need to know how do you do them.
I missed class, and our textbook is so bad that it might as well be written in a foreign language. I understand how to do dy/dx of an equation not in the form y =... ex. y^2 = x^3 + 2x + 5, (y'=(3x^2 +2)/2y) for example, but how would you take the derivative with respect to a variable not in the equation, t for example, and how would you take dx/dy for an equation?

Are you talking about y being a function of a single variable or more than one variable?

If you are thinking of y being a function of more than one variable, you are talking about the partial derivative and if you have, say, $y^2= x^3+ 2x+ 5$, and you want to differentiate with respect to t, with x and t independent variables, then $2yy_t= 0$ so that $y_t= 0$. That is, since y, according to your equation, does NOT depend on t, they the derivative of y with respect to t is 0.

The same thing, of course, happens if you are thinking of y as a function of the single variable x. If you are given that, say, $sin(y)+ y^2= e^y$, so that x does not appear in the equation, you can go ahead and differentiate with respect to x: \(\displaystyle cos(y)y'+ 2yy'= e^y y'[/itex] but that all reduces to $y'(cos(y)+ 2y- e^y)= 0$ so that $y'= 0$. Again, the fact that the independent variable does not appear in the equation means that y does NOT, in fact, depende upon that variable and so the derivative is 0.

Now, about the derivative of x with respect to y. If you have, say, $y^2 = x^3 + 2x + 5$ you can simply differentiate each term in the equation with respect to y, using the chain rule for differentiating functions of y: $2y= 3x^2 dx/dy+ 2 dx/dy= (3x^2+ 2)dx/dy$ so that
$$\frac{dx}{dy}= \frac{2y}{3x^2+ 2}$$

Or you can use the basic rule
$$\frac{dx}{dy}= \frac{1}{\frac{dy}{dx}}$$

As you said before, if $y^2= x^3+ 2x+ 5$ then $2yy'= 3x^2+ 2$ so that
$$y'= \frac{2y}{3x^2+ 2}$$
and so
$$\frac{dx}{dy}= \frac{1}{y'}= \frac{2y}{3x^ 2+ 2}$$
as before.\)

 

[Deathcrush]

why don't you just check another book?

 

[omenka]

The routine is simple, perform your usual differentiation, noting that y(x) is a function of x and so must use chain rule, factor out and solve for dy/dx. If your book is lousy, check out the earlier editions of Calculus and Analytic Geometry by G. Thomas, the book I used in high school in the old country. Implicit differentiation is great for finding derivatives of complicated functions where it may be formidable or impossible to solve for y as a function of x before performing your differentiation. To test if y could be solved in terms of x, use Abel's theorem. Send me a mail if you need additional help.

Henri Onuigbo

 

[omenka]

Things you should know about differential equations in general and derivatives in particular.

Separable equations: In general, an ODE is separable if it can be put in the form Recognizing where an ODE is separable is usually, but not always easy. If the equation appears in the form If the function is not so factorable, then the ODE is not separable. Alternatively, when an equation appears in the form We can try to factor both functions and reduce it to the type above. Now, how do we test for separability in general? To test whether a given function F(x, y) has a factorization of the form f(x)g(y), first select initial values of x and y such that , always use (0,0). If (1) holds, then the ODE is separable, otherwise forget it. Linear and Rational Substitutions In certain situations, ODE could be rendered separable by two types of substitutions: Consider another example Either linear or rational substitution applies when a given ODE can be put in the form The rational substitution applies to ODE of the form: Now we explain why the linear substitution works: This can be regarded as a formula whenever a linear substitution applies. Let me also show you why rational substitution works Again regard this as a formula whenever rational substitution applies. How do we know when to apply which substitution? For the equation in the form in which the coefficient functions are polynomials in x and y such that every term in both polynomials has the same total degree. In this situation rational substitution must apply. Take the example: The degree of the polynomials need not be integers. For example Another powerful substitution could be used whenever the ODE is of the form: We can therefore use the following formula to solve the non-separable ODE whenever it is of the form above: Use trig substitution to complete the job. Exact Equations Partial derivatives of a function of two variables F(x, y) are calculated by treating one variable as a constant and differentiating with respect to the other. A differential equation of the form Can be written as In the case above, the solution to the DE is simply Such a function is guaranteed to exist iff . In this case the DE is called exact. In the event, the DE is not exact, then we need an integrating factor: Given the equation and suppose that Then the integrating factor for the DE is given by Multiplying the DE by the integrating factor renders the DE exact. Similarly, given the DE , and suppose Then the integrating factor for the DE is given by More generally, suppose we want to search for an integrating factor of the special form for the DE . We must arrive at the conclusion that The problem is to determine values of m and n that satisfies the above. Consider the DE Multiplying the DE by the above IF renders it exact. We can also extract a separable integrating factor of the form This last equation includes (1a) and (1b) where Q(y) = 0 or P(x) = 0, as well as (2) as a special case. It will be wrong, however, to say that the last equation covers all possible integrating factors. First Order Linear Equations We can simplify things by making using equation (2) above, in that case There are three terms in a linear equation, whenever one of the terms is missing, the resulting linear DE is separable, and that's the law. Let a and b be constants and An equation of the form has a solution where r is a constant. Furthermore, an equation of the form has a solution of the form How do we find r? Substitute in and solve for r, and the final solution is . Now how do we solve the general nonhomogeneous linear ODE? Easy: This brings us to the alternative to the integrating factor method: we simply solve two separable equations. Given the linear ODE (1) we first find one nontrivial solution of the RHE Because the RHE is separable, this is very easy. (2) Then substitute , into the original ODE. The result is an ODE in v and x which is always separable. There is a very direct solution of linear nonhomogeneous ODE for those of us that love formulas: but we shall first restate the ODE in the form The ODE , is prevented from beinng linear by the second degree term in y. Suppose this factor is divided out, we have The nonlinear ODE is called a Bernoulli DE. And the general Bernoulli DE is always in the form Where n can be any real number, of course if n is either 0 or 1, then the DE reduces to linear equation. Hence the procedure for solving the Bernoulli equation is this. Divide the BDE by and substitute. ) This is a linear equation in v as the dependent variable. The Riccati equation is a class of equations that can sometimes be reduced to Bernoulli equation. The difficulty in solving RDE is that we must know one solution before we can obtain the complete solution. . Now consider the example The result is a Bernoulli equation with n = 2. In general, given the Riccati equation , let , be one known solution, then the substitution , converts the RDE to a Bernoulli equation with n = 2, and the resulting Bernoulli equation is Consider the following ODE , this equation cannot be solved by any of the methods of this section. It is a Riccati equation for which there is no solution available to convert it to a Bernoulli equation, so everything breaks down. The two-step method of linear equations also applies to Bernoulli equations: (1) Solve the RHE of the Bernoulli (2) Let this solution be (3) Substitute Into the Bernoulli equation to obtain Clearly, this is a separable equation; solving for v and using the relation , produces the solution to the Bernoulli equation. Examples of Homogeneous Stuff Take a few more examples: Answers; Examples of Some Linear Stuff Check for exactness and Solve Solve by finding a suitable integrating factor For those of us that love formula, let us state one for solving a general Bernoulli equation. These are some examples from qualifying examination at WSU. We can directly use the formula for linear equations. Clearly, for n = 0, we recover the formula for linear equations. Now let us develop the two step process for solving LDE. Let be the solution of the RHE Now substitute , into the LDE
Henri Onuigbo, P.E.

 

[omenka]

I see that the mathematics did not print in the above thread. I don't know how to fix it to accommodate the mathematics. I wrote the piece in MS Word and imported it to this Forum, but it did not translate. Any suggestions from the computer people ?

Henri Onuigbo (Omereada Nke Mbu)