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Question 9
i Chrome File Edit View History Bookmarks People Window Help €: * 9896L%), Sun Jun 25 1:14 PM Q E Hone - Ellucian - Luminis Pla x Files × e G square root symbol . Google × Chegg Study Guided Solutio Secure https://canvas.pasadena.edu/courses/1078108/files?preview-61462553 | AMPLE EXAM 1.pdf Download Info Close ipl...rk -ZOOM + answers not transferred to the answer sheet. y=cos( Le 2. Differentiate the function. y cos . Evaluate the integral. 5. Use logarithmic differentiation to find the derivative of the function. y- (ln)N 7. Find the derivative of the function. Simplify where possible. y-xsin"k + 8. + e 6. Evaluate the d , Evaluate the integral. Evaluate the integral. J. 9, Findy, for tan-try=x+xy2 10. Evaluate the integral. d 11. Find the limit. lim (4x 12. Ifx) is continuous,2)0, and (2) 13. Find the limit lim 2 +3x) +f2+5x) 7, evaluate lim (hint: Use Hospital's Rule) Find the limit lin Find tan-1(tanh x) When a cold drink is taken from a refrigerator, its temperature is 5°C. After 25 minutes in a 20°C room its temperature has increased to 10'C. 14. 15. a. What is the temperature of the drink after 50 minutes? b. When will its temperature be 15'C? 25Explanation / Answer
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When we use implicit differentiation, we derive both sides of the equation using the chain rule. I will use the notation y' to indicate derivative.
Recall the derivative of tan-1(x) = 1 / (1 + x2)
d/dx[tan-1(x2y)] = d/dx[x + xy2]
(2xy + x2y') / (1 + x4y2) = 1 + y2 + 2xyy'
Multiply both sides of the equation by the left side's denominator .
2xy + x2y' = (1 + x4y2)(1 + y2 + 2xyy')
2xy + x2y' = 1(1 + y2 + 2xyy') + x4y2 (1 + y2 + 2xyy')
2xy + x2y' = 1 + y2 + 2xyy' + x4y2 + x4y4 + 2x5y3y'
Now that the hard part is done, we move all the y' terms to the left side of equation and all the non-y' terms to the right side of equation.
x2y' - 2xyy' - 2x5y3y' = 1 - 2xy + y2 + x4y2 + x4y4
Factor out the y' on the left side of equation.
y' (x2 - 2xy - 2x5y3) = 1 - 2xy + y2 + x4y2 + x4y4
Finally, divide both sides of the equation by the coefficient of y'.
y' = ( 1 - 2xy + y2 + x4y2 + x4y4 ) / (x2 - 2xy - 2x5y3)