1. For f(x) = e^x^2+cos(x), calculate f\'(0) by using the definition and then by
ID: 2894155 • Letter: 1
Question
1. For f(x) = e^x^2+cos(x), calculate f'(0) by using the definition and then by using the chain rule! 2. Find f'(4) directly from the definition of the derivative given f(x) = squareroot x - 1. 3. Find h'{x) given that h(x) = squareroot x^2 + 1 4. Assuming that the equation determines a function f such that y'= f(x), find y': 5x^2-2x^2y^2+4y^3-7 = 0. 5. Assuming that the equation determines a function/such that y = f(x), find y': squareroot x+1/squareroot y+1= y^2. 6. Find an equation of the tangent line to the graph y = (x^3 4- 2)^5 at x = -1. 7. Find an equation of the tangent line to y = x^4 + 2x^2 - x at x = 2. 8. Find the critical numbers of g(t) = t^2/5t+4 9. Find the maximum volume of a right circular cylinder that can be inscribed in a cone of altitude 12 cm and base radius 4 cm. if the axis of the cylinder and cone coincide. 10. Evaluate integral 4 sin(x) Practice dxExplanation / Answer
( 10 )
4 sin(x) dx
===> 4 sin(x) dx
===> 4 ( - cosx ) + c
==> - 4 cosx + c
( 8 )
given g(t) = t^2 / 5t + 4
g'(t) = d/dx ( t^2 / 5t+4)
==> 2t ( 5t+4) - t^2 ( 5) / ( 5t+4)^2
==> 10t^2 + 8t - 5 t^2 / ( 5 t + 4 )^2
g'(t) = 0
==>10t^2 + 8t - 5 t^2 / ( 5 t + 4 )^2 = 0
==> 5 t^2 + 8 t = 0
==> t ( 5 t + 8 ) = 0
t = 0 , 5 t + 8 = 0
t = 0 , - 8 / 5 are critical numbers
( 7 )
given y = x^4 + 2 x^2 - x at x = 2
y' = slope m = d/dx ( x^4 + 2 x^2 - x )
==> 4 x^3 + 4 x - 1
at x =2
y' = 4(2)^3 + 4(2) - 1 ==> 32 + 8 - 1 ==> 39
y = ( 2)^4 + 2 (2)^3 - 2 ==> 16 + 16 - 2 ===> 30
equation of tangent line is
y - 30 = ( 39) ( x - 2)
y = 39 x - 78 + 30
y = 39 x - 48
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( 6 ) given y = ( x^3 + 2 )^5, at x = -1
y' = slope m = d/dx ( ( x^3 + 2 )^5 ) ==> 5 ( x^3 + 2 )^4 d/dx( x^3 + 2 )
==>5 ( x^3 + 2)^4 ( 3 x^2) ==> 15 x^2 ( x^3 + 2)^4
at x =-1
y' = 15 (-1)^2 ( (-1)^3 + 2)^4==> 15( -1 + 2) ^4==> 15
y = ( (-1)^3 + 2 )^5==> ( - 1 + 2 )^5 ===> 1
equation of tangent line is
y - 1 = ( 15) ( x + 1)
y = 15 x + 15 + 1
y = 15 x + 16
( 3 )
h(x) = sqrt( x^2 + 1)
h(x) = ( x^2 + 1)^1/2
h'(x) ==> d/dx ( ( x^2 + 1)^1/2 )
==>( 1/2 )( ( x^2 + 1)^1/2-1 ) d/dx( x^2 + 1)
==> ( 1/2 )( ( x^2 + 1)^-1/2 ) ( 2x)
==> x/ ( x^2 + 1)^1/2
( 2 ) given
f(x) = sqrt ( x - 1)
f'(x) ==> d/dx ( ( x - 1)^1/2 )
==>( 1/2 )( ( x - 1)^1/2-1 ) d/dx( x - 1)
==> ( 1/2 )( ( x^2 + 1)^-1/2 ) (1 )
==> 1/ 2( x - 1)^1/2
f'(4) ==> 1/2(4-1)^1/2==> 1/2 ( 3)^1/2==> 0.288675
( 1 )
given
f(x) = e^x^2 + cosx
let u = x^2 + cosx
f'(u) = d/du ( e^u ) d/dx ( x^2 + cosx)
==> e^u ( 2x - sinx)
f'(x)==> e^( x^2 + cosx) * ( 2 x - sinx )
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