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XLequation has (1,when 3ejoqejed the tangentline tangent whose Practice Another Version SCalcET8 3.11075 ZXe the Ilne "wth equation = SCalcET8 3 1.509 XP i Dif...

Question

XLequation has (1,when 3ejoqejed the tangentline tangent whose Practice Another Version SCalcET8 3.11075 ZXe the Ilne "wth equation = SCalcET8 3 1.509 XP i Differentiate the functicn Need Help? II 4 point? what Need Help? the 1 1 Find For

XL equation has (1, when 3 ejoqejed the tangent line tangent whose Practice Another Version SCalcET8 3.11075 ZXe the Ilne "wth equation = SCalcET8 3 1.509 XP i Differentiate the functicn Need Help? II 4 point? what Need Help? the 1 1 Find For



Answers

Find an equation of the tangent line at the indicated point or number. $$ \frac{1}{x}+\frac{1}{y}=1 ; \quad x=3 $$

And this problem will be finding the equation of a tangent line to our function at a given point. So first to find that equation of the tangent line only to find the slope. Then we'll use that given point to find the entire equation will be also using this difference quotient that you have seen in previous problems. That's thinking of the tangent as a limit. As H approaches zero, we evaluate our function at the given value plus each and then we subtract our function evaluated at a are given value. So this comes from the slope formula and again this difference quotient helps us find the slope of our tangent line. So what A. Is is X. Value of our function? Our function in this problem is why is equal to x squared -3 x plus one. So we can go ahead and evaluate our function for a and also a plus H. So evaluated at a yes of three, that would be equal to three squared -3 times three plus one. If we simplify we have nine -9 Plus one. The nines cancel out. So we're just left with one. Now we can evaluate the function for F. of three plus H. Substituting in three plus H. Everywhere there is an ex so three plus H squared minus three Times three Plus H. That's what first thing we'll do is expand that binomial expression 3-plus h. multiplying it by itself besieged Times three Plus Age. I'll also take the step here of distributing that -3. We have negative three times three is negative nine and negative three Times H. Would be -3 H. His plus one. So let's multiply together. What we initially expanded three times three would give us 93 times eight would give us three age. Each time three would give us three h. And each times age would give us each square. Also want to add that -9 -3 each and plus one. Now we can start combining like terms, see some several terms with an H. And several constants that we can combined. So those nines right away, we can see those cancel out. We're just left with a constant of one. Also three H&M -3 H cancel. So the simplified expression would be three H plus H squared plus one and we can take that what we just found, substitute it back into our difference quotient. So we're thinking of that tangent slope as a limit. When H approaches zero We have three h plus H squared plus one and then minus what all over. H. Those ones cancel. And we're left with two terms in our numerator that both have an H. In them. So we can factor to each from the three, three H would just give us three and we can factor out of each from each square would just give us a church. We can divide that common factor of H. And our simplified expression that we're left with could be three plus H. And now we're ready to evaluate that limit Subsequuting in zero for where ages. So the limit as H approaches zero, we have three passage would just be three. And we can think of that as the slope of the tangent line. So now that we have the slope we can start creating an equation and I'm going to choose creating an equation in slope. Why intercept form or why sequel the mx plus B. You know, MS three. So we can substitute that in for em and we know a point on our line earlier We had the .31 so we can substitute in one for why And three for X. It's one of the points on our line and or tangent line and then we can solve and simplify. So we have one is equal 2 9 Plus B. And as we saw for B and subtract nine from both sides. We have negative eight is equal to be. So let's rewrite that slope intercept equation Now that we have the Y intercept, we have y is equal to three X -8. And we could even go with one step farther if we wanted to write this in a different form, that's equal to zero. We want zero on one side of our equation and subtract Y from both sides. So we have zero is equal to three x minus y minus eight. The equation of that tangent line.

And the problem we have two by three extra. The power To buy 3 -1 -2.3. Why do the power to upon 3 -1? Why does -2? I does equal zero. So this is to upon three Extra. The power -1 upon three minus two upon three. What do the power minus one upon three? Why does that -2? What does that equal? 0? So to upon three Extra par -1 up on three equal what it does two plus two upon three. Why did the par -1 up on 3? So why does become two upon three Extra Power -1 upon three bond two plus two upon three. Why did the power -1 upon 3? Now we have why? That's at one and -1 that equals two. Uh huh. Yeah four. The equation of tangent becomes Why- of -1 that equals half X minus one. So it is Y plus one equals a half x minus one. So this is our answer.

In this question we are given the curve X over Y plus y. Over x cubed is equal to two and we have to find detergent to this curve at the point minus one minus one. Now we know that to find the gradient of this tangent. We had to find to I. D. X. And here we are going to apply implicit differentiation but first let's simplify this as over Y plus. Here we have Y cubed over X cubed Is equal to two. Multiplying by the lowest common denominator to eliminate nominators. Here we we have our laws comment. Nominator will be why execute multiplying everything in the equation with our lowest common denominator we have X to the power of four plus. Here we have Why to the power of four is equal to two. Cute. And now applying implicit differentiation You get four x to the power of three Plus four Y to the power of three. Do I. D X is equal to three X squared by two. I we held y constant here plus and now holding execute constant, we get to do I dx. Now taking due ideas terms to the left, We have four. White cubed minus two. X cubed is equal to Here. We have six x squared y -4 X. Cute. And therefore our do I. D X will be equal to six x squared y -4 x cubed over four, y cubed minus to execute. Now plugging in our .1 -1. We have six minus one squared minus one -4 and here we get -1 over four We Get -1, Cute -2. Right this way On the numerator there we have -6 and plus for over On the denominator we have -4 plus two, That's 2 -4. Class two. And here we get -2 Over -2 which gives us yeah gradient of positive one. Therefore the equation of our tangent will be Y is equal to x plus C. Now plugging in our point again to find the value of C we have minus one is equal to minus one plus C and get our C is one minus one, It is equal to zero. Therefore the equation of our tangent at a given point is God why is he going to X? And this is our final solution. That's it.

So here we give an example of specific cubic function, let's say it has this sort of a shape and were given information that we're interested in the tangent line at the point. Let's call the point here 2:03. So the tangent line would be why I minus the value of the function at a certain point. So why minus Y. Of two would be equivalent to the derivative of the function at the X value times x minus The certain x value or interest in which the X -2. So this will give the equation of the tangent line. So we have function Y equals execute minus three X plus one. And if we take the derivative of this this would be equivalent to three X squared minus three. If we evaluate the derivative at the point X equals two, So be four times maybe four times 3, 12, 12 -3 is nine. So we'll be getting information. This would be rewriting this would be AY -3 Equals nine times x -2. And this means that Y equals nine x minus 18 plus three, which B nine x minus 15. And this is our final answer.


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