Video Transcript
If 𝑓 of 𝑥 is equal to three 𝑥 sec 𝑥 plus two 𝑥 squared sin 𝑥 plus 11𝜋, find 𝑓 prime of 𝑥.
In order to find 𝑓 prime of 𝑥, we simply need to differentiate 𝑓 with respect to 𝑥. We can use the sum rule of differentiation in order to split up this differential. Now, the last term here is the differential of 11𝜋 with respect to 𝑥. However, 11𝜋 is a constant. Therefore, this differential we will evaluate to zero.
Now, for the other two differentials which we need to calculate, we’re differentiating products. In the first term, we have the product of three 𝑥 and sec 𝑥. And in the second term, we have the product of two 𝑥 squared and sin 𝑥. In order to differentiate these, we’ll need to use the product rule. The product rule tells us that if we’re trying to differentiate the product of two functions of 𝑥, for example, 𝑢 and 𝑣, then it’ll be equal to 𝑢 timesed by d𝑣 by d𝑥 plus 𝑣 times d𝑢 by d𝑥.
Let’s start by considering the differential of three 𝑥 sec 𝑥. In this case, we can set 𝑢 equal to three 𝑥 and 𝑣 equal to sec 𝑥. Now, we need to find d𝑢 by d𝑥 and d𝑣 by d𝑥. In order to differentiate three 𝑥 with respect to 𝑥, we need to use the power rule for differentiation. We multiply by the power which is one and then decrease the power by one. This gives us that d𝑢 by d𝑥 is equal to three.
Next, we need to find the differential of sec 𝑥 with respect to 𝑥. Now you may already know what this differential is equal to. However, we can work it out using the quotient rule. The quotient rule tells us that if we’re differentiating the quotient of function of 𝑥, so 𝑠 over 𝑡 with respect to 𝑥, then it will be equal to 𝑡 d𝑠 by d𝑥 minus 𝑠 d𝑡 by d𝑥 all over 𝑡 squared. Now, in our case, the reason why we can use the quotient rule here is because sec 𝑥 is actually equal to one over cos 𝑥, where 𝑠 is equal to one and 𝑡 is equal to cos 𝑥.
Now, we need to find d𝑠 by d𝑥 and d𝑡 by d𝑥. 𝑠 is equal to one which is a constant. Therefore, d𝑠 by d𝑥 must be equal to zero. And we can find d𝑡 by d𝑥 by simply differentiating cos of 𝑥. This gives us negative sin of 𝑥. Now, we can substitute these values back into the quotient rule to help us evaluate d𝑣 by d𝑥. And now, we can simplify this to see that it’s equal to sin 𝑥 over cos squared 𝑥. And we can separate out one over cos 𝑥 to enable us to say that d𝑣 by d𝑥 is equal to tan 𝑥 times sec 𝑥.
Okay, so now we know the values of 𝑢, 𝑣, d𝑢 by d𝑥, and d𝑣 by d𝑥, we’re ready to apply the product rule. We’ll obtain that the differential of three 𝑥 sec 𝑥 with respect to 𝑥 is equal to three 𝑥 tan 𝑥 sec 𝑥 plus three times sec 𝑥. Now, we just need to find the differential of two 𝑥 squared sin 𝑥. We can again use the product rule. This time, we’ll have that 𝑢 is equal to two 𝑥 squared and 𝑣 is equal to sin 𝑥. We need to find d𝑢 by d𝑥 and d𝑣 by d𝑥.
In order to differentiate 𝑢 with respect to 𝑥, we multiply by the power and decrease the power by one, giving us that d𝑢 by d𝑥 is equal to four 𝑥. Next, we can differentiate sin 𝑥 with respect to 𝑥. And this simply gives us cos 𝑥. Now, we substitute the values of 𝑢, 𝑣, d𝑢 by d𝑥, and d𝑣 by d𝑥 into the product rule to obtain that d by d𝑥 of two 𝑥 squared sin 𝑥 is equal to two 𝑥 squared cos 𝑥 plus four 𝑥 sin 𝑥.
Now, we have found all the components we need in order to find 𝑓 prime of 𝑥. This gives us the solution that 𝑓 Prime of 𝑥 is equal to three 𝑥 tan 𝑥 sec 𝑥 plus three sec 𝑥 plus two 𝑥 squared cos 𝑥 plus four 𝑥 sin 𝑥.
