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3.6, #8, Given Problem: 
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Here we integrate implicityly, remembering that whenever we encounter a "y", we must use the chain rule. The product rule was used on the right side. |
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Simplifying the both sides we get.... |
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Now we get the dy/dx terms on the same side of the equation. |
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And we factor out the dy/dx because that's what we need to solve for. |
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We solve for dy/dx by dividing both sides by the quantity on the left. |
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Factoring the numerator and denominator gives us.... |
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And finally, we have our implicit derivative. |
3.6, #18, Given Problem.
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Find an equation of the tangent line to the curve at the given point:
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Given problem. Number 18, lesson 3.6. |
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We begin by implicitly differentiating both sides of the equation. We use the product rule on both sides and the chain rule. |
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In the first line of this step, y' is factored out of the two quantities on the right, and in the second line, we factor the y' out of the entire right side. |
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In this step we get the y' terms on the left side and non y' terms on the right. |
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Now we factor y' out of the entire left side. |
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We now can solve for y' by dividing both sides by the large quantity which was multiplied by y' in the last step. |
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If we factor a 2 out of the numerator and denominator, and cancel, we get this. |
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when x = 0 and y = -2, y' = 0 so, the equation of the tangent line is: y = -2 |
Here we substitute the given values for x and y and find that the slope of the tangent line at the given point is zero. Then we can find the equation of the tangent line in slope-intercept form. |
3.6, #34, Given Problem.
Find the derivative of:  |
Given problem. Number 28, Lesson 3.6 |
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We use the power rule and begin to apply the chain rule to the inverse sine of x. |
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Here we finish the chain rule, using our newfound derivative of the inverse sine function. |
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The result is algebraically simplified. |
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