d/dx 1/x
-1/x2
Find the slope of the graph of the function at the given point.
f(x)=2x4-8 at x=1
f'(1)=8
d/dx 2/3x3
y'=-2/x4
Write the equation of the tangent line.
f(x)=3x2-4x at x=1
y+1=2(x-1)
d/dx 1/x1/2
y'=-1/2x3/2
Differentiate y=x³lnx
y′ =x²+3x2lnx
y=sin (x)/x4
y'=[x4cos(x)-4x3sin(x)]/x8
Find the equation of the normal line.
f(x)=8x-2+3x2 at x=-2
y-14=1/10(x+2)
d/dx x1/2-x-1/2
y'=1/2x1/2+1/2x3/2
Differentiate y=2x2cos(x)
y'=4xcos(x)-2x2sin(x)
f(x)= (x²-1)/ (x²+1), what is f′(x)?
f'(x)=4x/(x2+1)2
d/dx sec (x)
y'=sec(x)tan(x)
Find the equation of the tangent line.
f(x)=3cos x - 2x at x=0
y-3=-5(x-0)
Differentiate y=14x2/7
y′ = 4/x5/7
Differentiate and simplify
y=(4x2-3)(2x3-1)
y′ =40x4-18x3-8x
If velocity is v(t), find the acceleration, a(1)
v(t)=90t/(4t+10)
a(1)=225/49
d/dx -cot(x)
y'= csc2(x)