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Differentiate each of the following. a) f(x)=(x3+5x+2)sin⁡(x)b) y=sin⁡(x)cos⁡(x)c) g(x)=3x1+tan⁡(x)d) tan⁡(x)−6x2+cos⁡(x) Let y=...

Question

Differentiate each of the following. a) f(x)=(x3+5x+2)sin⁡(x)b) y=sin⁡(x)cos⁡(x)c) g(x)=3x1+tan⁡(x)d) tan⁡(x)−6x2+cos⁡(x) Let y=tan⁡(x)−1sec⁡(x).Show that y′=1+tan⁡(x)sec⁡(x).Find all x-values wherethe graph of f(x)=x−2cos⁡(x)hasa horizontal tangent line.

Differentiate each of the following. a) f(x)=(x3+5x+2)sin⁡(x) b) y=sin⁡(x)cos⁡(x) c) g(x)=3x1+tan⁡(x) d) tan⁡(x)−6x2+cos⁡(x) Let y=tan⁡(x)−1sec⁡(x). Show that y′=1+tan⁡(x)sec⁡(x). Find all x-values where the graph of f(x)=x−2cos⁡(x)has a horizontal tangent line.



Answers

Lines tangent to parabolas a. Find the derivative function $f^{\prime}$ for the following functions $f$. b. Find an equation of the line tangent to the graph of $f$ at $(a, f(a))$ for the given value of $a$. c. Graph $f$ and the tangent line. $$f(x)=3 x^{2} ; a=0$$

Using the curve tangent of X y equals X plus y. We're going to be finding the first derivative of this curve, and then we're going to be using that first derivative equation to find the slope of the curve at this 0.0 at the origin. So first off, we're going to have to find the derivative with respect to X of each of the terms in the original equation. So the derivative of Tangent of X y equals the derivative with respect to X of X close the derivative with respect to X of Why so on the left side of the equation, we're going to need to use the chain rule on the product rule. So using the chain drill, we know that the derivative of tangent is going to be C can't squared. So see, can't squared of that inner term x y times the derivative of that inner term x y. And for that, we're going to need to use the product rule So we'll see ex prime times why plus next times wide prime on the right side of the equation, the derivative with destructor X of X is simply one, and the derivative of why is derogative off? Why with respect to X and that is the term that we're going to be looking for to solve for the slope. So now we're going to rewrite the left hand side of this equation. So we have c can't squared of X Y times. The derivative of X is one. So we have why, plus x times, the derivative of why equals one plus the derivative term. So we want all of the derivative terms to be on one side and all of the non derivative terms to be on another side. So we're going to need to do this algebraic Lee. We're going to need to multiply through on the left side with that see, can't squared of X y So we'll have c can't squared of X y times Why plus x times the derivative times see can't squared of X y cools one plus the derivative. So we have the derivative terms here and here and the non derivative terms here and here. So you want them all to be on the same side. So we're going to subtract by the derivative on both sides and by C can't squared of X y times. Why on both sides. So that's going to give us on the left side of the equation. X times the derivative times See Can't squared of X Y minus. The derivative equals one minus. Why? See? Can't squared of X Y on the right side. So now we're going to factor out the derivative on the left side. So we have the derivative times X c can't squared of X Y minus one equals one minus Y c can't squared of X y So now we're going to The final step is going to be to divide both sides by X he can't squared of X Y minus one so that our derivative equation will be one minus. Why, c can't squared ec of x y over x c can't squared of X Y minus one. So now that we have this first derivative term, we want to plug in thes zero and zero for the X and Y so that we could find the slope at this exact point. So we're going to find the derivative or the slope equals one minus zero times. See? Can't squared of zero times zero over zero times. See can't squared of zero time zero minus one. So the zero's air going to multiply to cancel out. So that will end up with the derivative equals one over negative one. In other words, the slope at the origin this.

Using the curve tangent of X y equals X plus y. We're going to be finding the first derivative of this curve, and then we're going to be using that first derivative equation to find the slope of the curve at this 0.0 at the origin. So first off, we're going to have to find the derivative with respect to X of each of the terms in the original equation. So the derivative of Tangent of X y equals the derivative with respect to X of X close the derivative with respect to X of Why so on the left side of the equation, we're going to need to use the chain rule on the product rule. So using the chain drill, we know that the derivative of tangent is going to be C can't squared. So see, can't squared of that inner term x y times the derivative of that inner term x y. And for that, we're going to need to use the product rule So we'll see ex prime times why plus next times wide prime on the right side of the equation, the derivative with destructor X of X is simply one, and the derivative of why is derogative off? Why with respect to X and that is the term that we're going to be looking for to solve for the slope. So now we're going to rewrite the left hand side of this equation. So we have c can't squared of X Y times. The derivative of X is one. So we have why, plus x times, the derivative of why equals one plus the derivative term. So we want all of the derivative terms to be on one side and all of the non derivative terms to be on another side. So we're going to need to do this algebraic Lee. We're going to need to multiply through on the left side with that see, can't squared of X y So we'll have c can't squared of X y times Why plus x times the derivative times see can't squared of X y cools one plus the derivative. So we have the derivative terms here and here and the non derivative terms here and here. So you want them all to be on the same side. So we're going to subtract by the derivative on both sides and by C can't squared of X y times. Why on both sides. So that's going to give us on the left side of the equation. X times the derivative times See Can't squared of X Y minus. The derivative equals one minus. Why? See? Can't squared of X Y on the right side. So now we're going to factor out the derivative on the left side. So we have the derivative times X c can't squared of X Y minus one equals one minus Y c can't squared of X y So now we're going to The final step is going to be to divide both sides by X he can't squared of X Y minus one so that our derivative equation will be one minus. Why, c can't squared ec of x y over x c can't squared of X Y minus one. So now that we have this first derivative term, we want to plug in thes zero and zero for the X and Y so that we could find the slope at this exact point. So we're going to find the derivative or the slope equals one minus zero times. See? Can't squared of zero times zero over zero times. See can't squared of zero time zero minus one. So the zero's air going to multiply to cancel out. So that will end up with the derivative equals one over negative one. In other words, the slope at the origin this.

We're going to start this by taking the derivative on both sides with respect to X. So we're gonna take DDX of tangent X y as a vehicle to the derivative of X with respect to or driven out of this respect to x both sides. So first, let's do this derivative over here. So the drive of this is going to be while the derivative tangent is seeking squared. So we're gonna have seeking squared X Y, and then we're going to multiply by the derivative off the inside. So that's gonna be the derivative of X. Y so right that now as this. So, for this one, we're going to need to use product rule here. Um, I'm gonna go ahead and be the right hand side, though, so this derivative is gonna be one plus and then derivative of why it's gonna be one times D y t x so times why prime? So it's gonna be here. We have what what's Yeah, one plus de y t x. Okay, so let's continue over here. This we're gonna need to use product rule, so we have second squared Next. Why? Times now we're going to have so first times derivative. Second means X times d y over the X and then plus second, um, so second times derivative of first. So that's just going to plus why, and that's evil to one plus de y over d X. Next, we're going to multiply this out here so we're gonna have so x de y the x or will have this Times X and then seek in squared X y and then plus why seek and squared X y that's equal to one plus de y t X. So we're going to move all of our d y t X terms to the other side and everything without a D by D. X to this side. So we'll have de y the x and then X seeking squared X y and then plus or sorry minus D y over the X And then this is gonna be able to one minus. Why? Seeking squared X Y Next. We're going to factor out on B y t X. So we're gonna have de y over d x times X seeking squared X Y minus one is equal to one minus. Why second squared X y Okay, I'll move all this over some here like that. So finally, we're going to get our final Ah, answer for D y t X here by dividing this to the site. So D y t x, this is gonna be equal to We have one minus. Why? Second squared X Y all over. Ex seeking squared X Y minus one. Now let's plug in our point to find de y the X at 00 So we get one minus zero times and then seek in squared of zero. Well, it doesn't matter because we have a zero up front here, so we get one minus zero, and then here, we're gonna have This is zero, um, times. And then doesn't matter what that is. That's gonna be zero minus one. So on the top, we get one on the bottom. We have negative one. So one divided by negative one is gonna be negative. One

So just like the previous tutorial. Uh I just want to take the implicit derivative in order to find the uh the equation of attention line. Right? So what is the derivative of this? Well, this is a function of a function. Okay, So you're gonna use the chain rule and the chain rose has to bring the dicks opponent down and then write the function again. And subtract one from the explaining explaining this to So when I subtract one is just gonna be one. So I'm not going to write it and then multiply by the derivative of the inside. So what is the derivative of this inside? Right here. Thing inside the parentheses, the derivative of that is just one. Okay, so I'm not going to write it. So this is the derivative. And then uh Plus okay then. Plus uh distributor of that one is going to be I bring to down and why? Plus four. I take one from the top. And that is one then multiply by the derivative of the inside. What is the derivative of this guy? Well I told you whenever you take the derivative of why you have to attach A. D. Y. D. X. So the derivative of this one is just gonna be do I the X. Right. I said that in the last two tour before this one. And then what is the derivative of the constant? Where that is zero? You want to sell for? Dy dx? So what you're gonna do is bring this one to the other side of the equation. So this is gonna be too uh in fact, I can just divide through by two, multiplying through by one half. So one half year, one half, year one half here, you can see that this one half is gonna take away this too. And this one is gonna take away this too. So whenever I multiply through by one half the twos are taken care of. Right? So all that I have is this that and in that now I want to take this one to the other side of the equation. So whenever I do that what is happening, I'm going to have y plus four Dy DX. And then this is gonna be negative X -3. Right? You subtract this one from the other side of the equation. You have this one. Now I want to divide both sides by this guy. So whenever I do that I have dy DX to be negative X -3. Okay And then in fact I wouldn't take away the negative. So what I can do is absorb this negative and just flip this one. Okay so this is gonna be three minus X. Okay whenever you foil you can see that this negative is gonna flip. this one gonna have three minus X. Okay And I'm gonna write the bible is had by this guy. I have uh you know y plus four. So this is what I have as the the yeah slope function. Now I'm gonna put the 0.0. Right? You want to find the slope at this 0.0 is whenever I put 00 here means that whenever I see X. I put zero whenever I see what I put zero. So whenever I do that you can see that do I. D. X. Right here? It's gonna be 3/4. Right? So the slope of the tangent line at this point It's the 3/4 right at this point, the slope of the tangent line at the point at the origin It's 3/4 right? So that is the slope of the tangent line. Now I want to find the equation. So I'm still going to use the same formula Y minus y. One equals m x minus x. One like we did in the last century before this one. So whenever I do that, you can have why minus? Well, why one is zero? Right? Because the point is 00 right here. So why one is zero? And then the slope is three quarters and then the X one is also zero. So finally you have y equals 3/4 or 3/4 times X. So this is the equation of the line with a tangent line. This is the slope, right? And now we want to sketch a little sketch right to uh uh display. So uh the Sketch is gonna be like, this is going to pass through the origin because the point is 00 right here, this is the origin. Okay, so whenever you have this crew titian coordinate system, the circle is going to be something like this, right? It's not drawn to scale right once again because my drawing is horrible, but I'm just listening to important points has to pass through the origin. And then the tangent line is gonna catch this curve at the origin, right? Remember the tangent line is just a straight line that touches the point, the curve or the circle at a particular point and this time is touching it at the origin, right? So that is this is a rough sketch of what is going on. Okay, And so this is the sketch, this is the tangent line right here and ah this is the slope of the of this tangent line. And this is the equation of the tangent line.


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