5-8. Use differentiation to verify each equation.

Question

d/dx (tan³ x-3 tan x+3x) = 3 tan⁴x

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says consider the equation below. Using differentiation, state if it is true or false, and we're given the derivative with respect to X of the quantity of 2 multiplied by cotangent of X multiplied by cos squad of x, minus 8 multiplied by cotangent of x minus 6 X is equal to minus 6 multiplied by cotangent to the 4th of X. We're given two possible choices as our answers. For choice A we have true, and for choice B, we have false. So we need to determine whether this statement is true or false by using differentiation. So we're going to be evaluating the derivative on the left-hand side of this equation, and we're going to do this term by term, since there are 3 terms in the quantity that we need to differentiate. So, looking at the first term, we're going to have the derivative. With respect to X, Of the quantity of 2 multiplied by cotangent of X, multiplied by cosequent squad of X. And in order to take this derivative, we're going to need to use our product rules. So we call for the product rule that if we take the derivative of the product of U multiplied by V, that is going to be equal to U multiplied by V plus U multiplied by V. So, that means our derivative here is going to be equal to and applying the product rule here. We'll have the derivative of the first term in a product, which is the derivative of 2 cotangent of X. I'm gonna take that derivative, I'm gonna get 2 multiplied by minus cosequent squared of X. That gets multiplied by the second term in our product, which is RV, and that is cosequence squared of X will multiply. That derivative by cosequence squared of X. Then we'll have plus our first term, which is 2 multiplied by cotangent of X, multiplied by the derivative of our second term, which is the derivative of cos squared of X. And so here we'll have to apply um our power rule and our chain rule. So we'll have 2 multiplied by cosequant of X. That gets multiplied by the derivative of cosequent, which is the quantity of minus cosequant of X multiplied by cotangent of X. We can simplify this expression, so this is going to be equal to. -2 multiplied by cosequent to the 4th of X. Minus 4 multiplied by cosequent squared of X multiplied by cotangent squared of X. So now we need to look at the second term in our derivative, and that it's going to be the derivative with respect to X of the quantity of 8 multiplied by cotangent of X. So this is going to be equal to 8, multiplied by the derivative of cotangent, which is minus cosequent squad of X. So this is just equal to -8, multiplied by cosequent squared of X. And then finally, for our last term, we need to take the derivative with respect to X. Of the quantity of 6 X. Which is just gonna be equal to sex. So now, if we look at the derivative, With respect to X, Of the quantity of 2 multiplied by cotangent of X, multiplied by coequant squared of X. 8 cotangent of X. Minus 6 X. That is going to be equal to. And here we'll just apply each of our um derivatives that we found, so it's gonna be minus. 2 multiplied by cosequent, the 4th of X minus or cosequent squad of X multiplied by cotangent squared of X. Plus Eat Cosequant squad of X. Minus 6. Now, what we want to do is write everything in terms of cotangent of X. So, this is going to be equal to. Minus 2, and here recall that cosequence square of X is equal to 1 plus cotangent square of X. So we'll have 2 multiplying the we'll have minus 2 multiplied by the quantity of 1 plus. Cotangent squared of X in quantity squared. -4, multiplied by the quantity of 1 plus cotangent squared of X. In quantity multiplied by cotangent squad of X. Then we'll have plus 8 multiplied by the quantity of 1 plus cotangent squared of X. In quantity minus 6. So now multiplying everything out, this is going to be equal to. -2. -4. Cotangent squad of X. -2. Cotangent to the 4th of X. -4. Cotangent squared of X. -4 cotangent to the 4th of X. Plus 8. Plus 8 cotangent squared of X minus 6. So now we just need to gather light terms, this is going to be equal to. So we'll start off by looking at terms that have cotangent to the 4th of X. So here we have minus 2 multiplied by cotangent to the 4th of X. Then we have minus 4 of multiplied by cotangent to the 4th of X, and when I add those together, I get -6 cotangent the 4th of X. Next, we'll look at terms that have cotangent squared of x in them, so we'll have minus 4, cotangent square of X, minus 4 cotangent squared of X, and then we'll have plus 8 cotangent squared of X, and those add up to 0. Then finally, we'll look at our constant terms, so we'll have minus 2. Plus 8 minus 6, which add up to 0. So that means that this derivative is equal to -6 multiplied by cotension to the 4th of X, and this is exactly what our statement said. And so that means that our statement is actually bro. So that is the answer that we were actually looking for. And so when we come back up to our answer choices and compare what we found to our answer to their answer choices, the correct answer choice corresponds to answer choice A. So I hope that this explanation has been useful, and I'll see you in the next video.

5-8. Use differentiation to verify each equation.d/dx (tan³ x-3 tan x+3x) = 3 tan⁴x

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5-8. Use differentiation to verify each equation.

d/dx (tan³ x-3 tan x+3x) = 3 tan⁴x