![]() We can generalize this by multiplying both sides of the equation by a constant □: Multiplying the constant of integration by a constant yields a constant, so we have We want to take out the factor of □ + 1 from the integral and divide both sides of the equation by □ + 1 however, Substituting these expressions into the reverse chain rule, we obtain We can recall that the power rule for differentiation tells us that □ ′ ( □ ) = ( □ + 1 ) □ . Let □ ( □ ) = □ for some unknown constant □ and let □ ( □ ) be aĭifferentiable function. There are many applications of the chain rule however, in this explainer, we will focus on two specific applications of this result. If □ is differentiable at □ and □ is differentiable at □ ( □ ), This is known as the reverse chain rule since it is found by reversing the chain rule by integration. This is now in the form of an integral result, where we need to add a constant of integration as usual: We can reverse this process by integrating both sides of this result with respect to □. Sense, a very useful application of this property is that any derivative result can be stated as an integration result by reversing the process.įor example, we recall that the chain rule tells us that if □ is differentiable at □ and □ isĭ d □ ( □ ( □ ( □ ) ) ) = □ ′ ( □ ) □ ′ ( □ ( □ ) ). Furthermore, we can verify our answer to an integration problem by differentiating it. This means that knowing how to perform differentiation on certain expressionsĬan help solve certain types of integration problems. Integration and differentiation are the reverse processes of each other. And we're done, and we could distribute this natural log of four if we found that interesting.In this explainer, we will learn how to evaluate integrals of functions in the form The natural log of four times X squared plus X. And so we can just rewrite this as two X plus one over over over the natural log of four. And then we're gonna multiply that times U prime of X. So this is one over the natural log of four. More steps just hopefully so it's clearer what I'm doing here. And of course, that whole thing times U prime of X. Times, instead of putting an X there it would be times U of X. So it's going to be one over the natural log of four. And you just do is you take the derivative of the green function with So, this is going to be equal to V prime of U X, U of X. If we want to know what V prime of U of X we would just replace wherever we see an X with a U of X. Well, what is V prime of U of X? We know what V prime of X is. This is going to be this is going to be the derivative of V with respect to U. And so we know from the chain rule the derivative Y with respect to X. And let me draw a little line here so that we don't get We have the whole expression that defines U of X. Remember, V is the logīase four of something. Information because Y this Y can be viewed as V of V of. You think of scaling the whole expression by one over the natural log of four. But we just scale it in the denominator with this natural log of four. Out of the change of base formulas that you might have seen. But since it's log base four and this comes straight If this was V of X, if V of X was just natural log of X, ourĭerivative would be one over X. ![]() So it's going to be one over one over log base four. Similar that if this was log base E, or natural log, except we're going to scale it. And then we've shown in other videos that V prime of X is, we're gonna be very ![]() We can say V of well if we said V of X this would be log base four of X. And we can say the log base four of this stuff well we could call that a function V. Derivative with respect to X of X is one. So that's gonna be I'm just gonna use the power rule here so two X plus one I brought that two out front and decremented the exponent. And it's gonna be useful later on to know what U prime of X is. So we could say we could say this thing in blue that's U of X. ![]() We're taking the logīase four, not just of X, but we're taking that What is the derivative of Y with respect to X going to be equal to? Now you might recognize immediately that this is a composite function. Let's say that Y is equal to log base four of X squared plus X. ![]()
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