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) =    

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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.

   

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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).    

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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 =    

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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) =    

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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).    

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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)=    

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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).    

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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).    

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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)=    

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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=