Chain rule
y=(2x+1)^2
y'=4(2x+1)
f(x)= x^2(2x+4)
Find f'(x)
f'(x)=2x(2x+4)+2x^2
f'(x)=4x(x+3)
Differentiate
y=1/(x+1)
y'=-1/(x+1)^2
Find the third derivative of
f(x)=x^3-x^2+x-1
f'''(x)=6x-2
Differentiate
4x^2+(4/x^2)
8x-8/x^3
Find y'(x) when
y=10(x^2-3)6
y'(x)=120x(x^2-3)^5
f(x)= e^(2x-2)(x^(-2))
Find f'(x)
f'(x)=2e^(2x-2)(x^(-2)-x^(-3))
Find f'(x) when
f(x)=(3x)/(x+5)
15/(x+5)^2
Find the fourth derivative of
f(x)=2x^3+x
f^4(x)=0
Find when the particle, x, is stationary if f(x)=x3.
Stationary at (0,0)
Find f'(x) when
f(x)=(5x^2-4)^7
f'(x)=70x(5x^2-4)^6
f(x)=x^3(2x+3)
f'(x)=?
f'(x)=8x^3+9x^2
Find the stationary points for
y=(2x+5)^2/x
Stationary points at
(-2.5,0) and (2.5, 40)
Find the fourth derivative of:
y=1/x
y^((4))=24/x^5
Find the equation of the tangent to the curve
y=x^2+5x-3
at the point (2, 11).
9x-y-7=0
Find f'(x) if
f(x)=(3x-1)^(1/2)
f'(x)= 3/2(3x-1)^(-1/2)
f(x)=(x+1)(2x+5)^4
f'(x)=?
f'(x)=(10x+13)(2x+5)^3
Differentiate:
(x^2+1)/e^(x^2-4)
-(2x^3)/(e^(x^2)-4)
Find the third derivative of
f(x)=x^(1/3)
f'(x)=(10x^(-8/3))/27
Find the equation of the tangent to the curve
y=x^2+2x-5
that is parallel to the line
y=4x-1
4x-y-6=0
f(x)=2(x^2-3)^5
Evaluate f'(2).
f'(2)=40
f(x)=e^(x^2-x^3)(50x^(1/2))
f'(x)=25(e^(x^2-x^3))(4x^(3/2)-6x^(5/2)+x^(-1/2))
Differentiate:
y=(\sqrt(x-1))/(2x-3)
y'=(-2x+1)/(2\sqrt(x-1)(2x-3)^2)
f(x)=e^(x^4)
f''(x)=?
f''(x)=2x^2e^(x^4)(6+8x^4)
A particle is moving along the x-axis. Its velocity, v, at time, t, is given by
v=\sqrt(20t-2t^2)
metres per second. Find the acceleration of the particle when t=4. Express your answer as an exact value in its simplest form.
a=\sqrt(3)/6 ms^(-1)