math
If f(x)=5ln(4+x)f(x)=5ln(4+x), f'(x)f′(x) = f'(2)f′(2) =
License Question 1. Points possible: 2 This is attempt 1 of 1.
If h(x)=6−3x3h(x)=6-3x3, find h'(3)h′(3). Use this to find the equation of the tangent line to the curve y=6−3x3y=6-3x3 at the point (3,−75)(3,-75). The equation of this tangent line can be written in the form y=mx+by=mx+b where mm is: and where bb is:
License Question 2. Points possible: 2 This is attempt 1 of 1.
Use the chain rule to find the derivative of
5e4x5−9x85e4x5-9x8
Use e^x for exex.
License Question 3. Points possible: 2 This is attempt 1 of 1.
Use implicit differentiation to determine dydxdydx given the equation x2+y5=eyx2+y5=ey. dydx=dydx=
License Question 4. Points possible: 2 This is attempt 1 of 1.
If f(x)=(4x2−4)(2x+2)f(x)=(4x2-4)(2x+2), Use the product rule to find: f'(x)f′(x) = f'(2)f′(2) =
Get help: Video
License Question 5. Points possible: 2 This is attempt 1 of 1.
If f(x)=4x+12f(x)=4x+12, find f'(x)f′(x).
License Question 6. Points possible: 2 This is attempt 1 of 1.
If f(x)=sin5xf(x)=sin5x, find f'(x)f′(x). Find f'(2)f′(2).
License Question 7. Points possible: 2 This is attempt 1 of 1.
Use the quotient rule to find the derivative of:
9ex+43x8−3x79ex+43x8-3x7
Use e^x for exex. You do not need to expand out your answer. Be careful with parentheses!
License Question 8. Points possible: 2 This is attempt 1 of 1.
If f(x)=13x+8f(x)=13x+8, find f'(−10)f′(-10).
License Question 9. Points possible: 2 This is attempt 1 of 1.
Use the product rule to find the derivative of
(8x6+2x4)(8ex−1)(8x6+2x4)(8ex-1)
Use e^x for exex. You do not need to expand out your answer.
Get help: Video
License Question 10. Points possible: 2 This is attempt 1 of 1.
The volume of a cylinder of height 10 inches and radius rr inches is given by the formula V=10πr2V=10πr2. Which is the correct expression for dVdtdVdt?
· dVdt=20πrdtdVdt=20πrdt
· dVdt=20πrdrdtdVdt=20πrdrdt
· dVdt=0dVdt=0
· dVdt=10πr2drdtdVdt=10πr2drdt
· dVdt=20πrdrdtdhdtdVdt=20πrdrdtdhdt
Suppose that the radius is expanding at a rate of 0.2 inches per second. How fast is the volume changing when the radius is 2.6 inches? Use at least 5 decimal places in your answer. answer: cubic inches per second
License Question 11. Points possible: 2 This is attempt 1 of 1.
If f(x)=6+8x+6x2f(x)=6+8x+6x2, find f'(x)f′(x). Find f'(1)f′(1).
Get help: Video
License Question 12. Points possible: 2 This is attempt 1 of 1.
Let f(x)=−6x(x−5)f(x)=-6x(x-5). Then f'(−1)=f′(-1)= . And after simplifying f'(x)=f′(x)= . Hint: You may want to expand and simplify the expression for f(x)f(x) first.
License Question 13. Points possible: 2 This is attempt 1 of 1.
Given f(t)=4et+6f(t)=4et+6 dfdtdfdt =
License Question 14. Points possible: 2 This is attempt 1 of 1.
If f(x)=tan3xf(x)=tan3x, find f'(x)f′(x). Find f'(3)f′(3).
License Question 15. Points possible: 2 This is attempt 1 of 1.
Find the slope of the tangent line to the curve −1x2+2xy−4y3=−11-1x2+2xy-4y3=-11 at the point (3,−1)(3,-1).
License Question 16. Points possible: 2 This is attempt 1 of 1.
If f(x)=−2x5−3x4−6x3x4f(x)=-2x5-3x4-6x3x4, find f'(x)f′(x).
License Question 17. Points possible: 2 This is attempt 1 of 1.
If f(x)=(4x2−4)(2x+5)f(x)=(4x2-4)(2x+5), find: f'(x)f′(x) = f'(1)f′(1) =
License Question 18. Points possible: 2 This is attempt 1 of 1.
If f(x)=ex11f(x)=ex11, find f'(x)f′(x).
License Question 19. Points possible: 2 This is attempt 1 of 1.
Let f(x)=x2+7x−11f(x)=x2+7x-11. Then f'(0)=f′(0)= . And after simplifying f'(x)=f′(x)= .
License Question 20. Points possible: 2 This is attempt 1 of 1.
Let f(x)=x3−2x+14f(x)=x3-2x+14. Then the equation of the tangent line to the graph of f(x)f(x) at the point (−4,−42)(-4,-42) is given by y=mx+by=mx+b for m=m= and b=b= .
License Question 21. Points possible: 2 This is attempt 1 of 1.
If f(t)=15t4f(t)=15t4, find f'(t)f′(t). [NOTE: Your answer should be a function in terms of the variable 't' and not a number! ]
License Question 22. Points possible: 2 This is attempt 1 of 1.
If f(x)=5f(x)=5, find f'(2)f′(2).
License Question 23. Points possible: 2 This is attempt 1 of 1.
Find the derivative of: (10−7x)2(10-7x)2. [Hint: expand first.] Now, find the equation of the tangent line to the curve at (1, 9). Write your answer in mx+bmx+b format y=y=
License Question 24. Points possible: 2 This is attempt 1 of 1.
Let u(x)=sin(x)u(x)=sin(x) and v(x)=x17v(x)=x17 and f(x)=u(x)v(x)f(x)=u(x)v(x). u'(x)u′(x) = v'(x)v′(x) = f'=u'v−uv'v2f′=u′v-uv′v2 =
Get help: Video
License Question 25. Points possible: 2 This is attempt 1 of 1.
If f(x)=4x8−6x5−6x3+6xf(x)=4x8-6x5-6x3+6x, find f'(x)f′(x).
License Question 26. Points possible: 2 This is attempt 1 of 1.
If f(x)=(x2+4x+8)3f(x)=(x2+4x+8)3, then f'(x)f′(x) = f'(5)f′(5) =
Get help: Video
License Question 27. Points possible: 2 This is attempt 1 of 1.
Find the derivative of: −2x4−6x6-2x4-6x6
License Question 28. Points possible: 2 This is attempt 1 of 1.
If f(x)=√10xf(x)=10x, find f'(x)f′(x). Find f'(2)f′(2).
License Question 29. Points possible: 2 This is attempt 1 of 1.
Find the derivative of: −9√x−10x4-9x-10x4. Type your answer without fractional or negative exponents. Use sqrt(x) for √xx.
License Question 30. Points possible: 2 This is attempt 1 of 1.
Find ddx(ln(6x+3))ddx(ln(6x+3))
License Question 31. Points possible: 2 This is attempt 1 of 1.
If f(x)=2x2−6x−5f(x)=2x2-6x-5, find f'(x)f′(x).
Get help: Video
License Question 32. Points possible: 2 This is attempt 1 of 1.
Use the chain rule to find the derivative of
10√10x8+5x61010x8+5x6
Type your answer without fractional or negative exponents. Use sqrt(x) for √xx.
License Question 33. Points possible: 2 This is attempt 1 of 1.
If f(t)=7t−6f(t)=7t-6, find f'(t)f′(t). Find f'(5)f′(5).
License Question 34. Points possible: 2 This is attempt 1 of 1.
If f(x)=2sec(5x)f(x)=2sec(5x), find f'(x)f′(x). Find f'(5)f′(5).
License Question 35. Points possible: 2 This is attempt 1 of 1.
Find the derivative of the function g(x)=(5x2+3x−2)exg(x)=(5x2+3x-2)ex g'(x)=g′(x)=
License Question 36. Points possible: 2 This is attempt 1 of 1.
Let f(x)=4x6√x+6x2√xf(x)=4x6x+6x2x. f'(x)=f′(x)=
Get help: Video
License Question 37. Points possible: 2 This is attempt 1 of 1.
Find the derivative of: −9√x−3x8-9x-3x8. Type your answer without fractional or negative exponents. Use sqrt(x) for √xx.
License Question 38. Points possible: 2 This is attempt 1 of 1.
Use the chain rule to find the derivative of
3(−7x8+8x5)193(-7x8+8x5)19
You do not need to expand out your answer.
License Question 39. Points possible: 2 This is attempt 1 of 1.
Use the chain rule to find the derivative of
4√7x3+6x547x3+6x5
Type your answer without fractional or negative exponents. Use sqrt(x) for √xx.
License Question 40. Points possible: 2 This is attempt 1 of 1.
If f(x)=3+6x−3x2f(x)=3+6x-3x2, find f'(−5)f′(-5).
Get help: Video
License Question 41. Points possible: 2 This is attempt 1 of 1.
If f(x)=1x2f(x)=1x2, find f'(3)f′(3).
License Question 42. Points possible: 2 This is attempt 1 of 1.
Find ddx(4ln(x))ddx(4ln(x))
License Question 43. Points possible: 2 This is attempt 1 of 1.
Find the equation of the tangent line to the curve y=5secx−10cosxy=5secx-10cosx at the point (π3,5)(π3,5). The equation of this tangent line can be written in the form y=mx+by=mx+b where mm is: and where bb is:
License Question 44. Points possible: 2 This is attempt 1 of 1.
Find ddx(45x+6)ddx(45x+6)
License Question 45. Points possible: 2 This is attempt 1 of 1.
If f(x)=(3x+7)−4f(x)=(3x+7)-4, find f'(x)f′(x). Find f'(2)f′(2).
Get help: Video
License Question 46. Points possible: 2 This is attempt 1 of 1.
If f(x)=6x+52x+4,f(x)=6x+52x+4,, find: f'(x)f′(x) = f'(2)f′(2) =
License Question 47. Points possible: 2 This is attempt 1 of 1.
If f(x)=cosx−3tanxf(x)=cosx-3tanx, then f'(x)=f′(x)= f'(3)=f′(3)=
Get help: Video
License Question 48. Points possible: 2 This is attempt 1 of 1.
Let f(t)=7t4−6t+5etf(t)=7t4-6t+5et. Then f'(t)=f′(t)=
License Question 49. Points possible: 2 This is attempt 1 of 1.
For what values of aa and bb is the line −4x+y=b-4x+y=b tangent to the curve y=ax2y=ax2 when x=−2x=-2? a=a= b=b=