![]() ![]() Differential calculus applies to anything that changes or moves over time. You have to use product rule, because it is the product of two functions.Calculus has massive applications to physics, chemistry, biology, economics and many other fields.The derivative attempts to extend the concept of slope to objects that are not lines.The problems in this exercise bring together several of the other rules, which is more common in real-life applications.Knowledge of all the differentiation rules, though particularly product and chain, are encouraged to ensure success on this exercise. ![]() The user is expected to answer the question and provide the answer in the space provided. Answer the conceptual question: This problem provides several potential functions and has a question that involves some application of the derivative rules.The user is expected to find the derivative of the function at the particular point. Find directly: This problem provides a composition of two functions and a particular point.There are two types of problems in this exercise: This exercise applies the chain rule in contexts where the product rule is also applied. It looks intimidating at first, but just say, okay, look.The Combining the product and chain rules exercise appears under the Differential calculus Math Mission. So this is going to be equal to negative 28 over positive four, which is equal to negative seven. Minus three times six, which is 18, all of that So you have five times negative two, which is negative 10, And then g prime of negative one, just circle it in this green color, g prime of negative F of negative one, so f of negative one, they tell us that right over there. G of negative one, well weįigured that right here. F prime of negative one is equal to five. We know, so first we wanna figure out f prime of negative one. So we actually know whatĪll of these values are now. Well negative one squared is just one, so this is going to be equal to six. So g prime of negative one is equal to six times Let's use the power rule,īring that three out front, three times two is six, x,ĭecrement that exponent, three minus one is two, and Power is just negative one, times two, so this is negative two, and g prime of x, I'llĭo it here, g prime of x. ![]() G of negative one is going to be two times negative one to the third power. We use the product rule when we need to find the derivative of the product of two functions - the first function times the derivative of the second, plus the. So, let's see, if this is, let's first evaluate g of negative one. They tell us f and fĪctually solve for those. One, what they are? Well some of them, they tell us outright. Now can we figure out whatį prime of negative one f of negative one, g of negative one, and g prime of negative ![]() All of that over, we'llĭo that in the same color, so take my color seriously. One times g of negative one, g of negative one minus f of negative one times g prime of negative one. It's going to be f prime of negative one, lowercase f prime, that'sĪ little confusing, lowercase f prime of negative Negative one is equal to, well everywhere we see an x, Let's just say we want to evaluate F prime when x is equal to negative one. To evaluate this thing, and you might say, wait, howĭo I evaluate this thing? Well, let's just try it. Prime of x, and likewise, you could say, well that is You could say this is the same thing as g Of writing this with a derivative operator, In the denominator, so times g of x, minus theįunction in the numerator, minus f of x, not taking its derivative, times the derivative in theįunction of the denominator, d, dx, g of x, all of that over, so all of this is going to be over the function in the denominator squared. Of the quotient rule, its derivative is going toīe the derivative of the function of the numerator, so d, dx, f of x, times the function In the denominator, we can say its derivative, and this is really just a restatement So if you have some function defined as some function in the numerator divided by some function Gonna state it right now, it could be useful to know it, but in case you ever forget it, you can derive it pretty quickly from the product rule, and if you know it, the chain rule combined, you can get the quotient rule pretty quick. So if we want to take it's derivative, you might say, well, maybe the quotient rule is important here. And the way they've set up capital F, this function definition, we can see that it is a quotient of two functions. So the way that we can do that is, let's just take theĭerivative of capital F, and then evaluate it at x equals one. Of x, and they want us to evaluate the derivative of capital F at x equals negative one. Let capital F be a function defined as, so capital F is definedĪs lowercase f of x divided by lowercase g Let g be the function g of x is equal to two x to the third power. A function such that f of negative one is equal to three, f prime of negative one is equal to five. ![]()
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