y=x^2+4sqrtx
y'=2x+2x^(-1/2)
y=tan(2x)
y'=2sec^2(2x)
y^2-x^2=4
dy/dx=x/y
y=x^(4x)
y'=x^(4x)(4+4lnx)
Find
(f^(-1))'(0) ... when ... f(x)=cscx-sqrt(2
-1/sqrt2
y=(2-x)/(3x+4)
y'=− 10/( 3 x + 4 )^2
y=csc^5(1-4x)
y'=20csc^5(1-4x)cot(1-4x)
y+cot^(-1)(y)+x-4=0
y'=(-y^2-1}/y^2
y=(secx)^x
y'=(secx)^x[ln(secx)+xtanx]
Find:
(f^(-1))'(-3) ... when ... f(x)=x^2+4x
1/2,-1/2
y= (x − 1 )^4 ( x + 1 )^3
y'=( x − 1 )^3 ( x + 1 )^ 2 ( 7 x + 1 )
y=arctan(x/2)
y'=2/(4+x^2)
4x^2-6xy^3+y=10
y'=(-8x+6y^3)/(1-18xy^2)
y=x^sqrtx
y'=x^sqrt(x)(lnx/(2sqrt(x))+1/sqrt(x))
Find the instantaneous rate of change of the function at x = 1.
f(x)=sin^2((pix)/4)
f'(1)=pi/4
y=xln(4x-1)^3
y'=12x/(4x-1)+ln(4x-1)^3
y=arcsin(e^(6x))
y'=(6e^(6x))/sqrt(1-e^(12x))
Find y''
y+siny=5x
(d^2y)/(dx^2)=(25siny)/(1+cosy)^3
y=((x+1)^4sqrt(x^2-5))/7^(2x-1)
y'=(4/(x+1)+x/(x^2-5)-2ln7)y
Find the x-value(s) where the tangent to the graph is parallel to the x-axis.
f(x)=3x^4-5x^3+2
x=0, 5/4
y=ln(e^(4)/(5x-1))
y'=-5/(5x-1)
y=cot(log(5^(2x-1)))
y'=-csc^2(log(5^(2x-1)))(2ln5)/(ln10)
Find y''
e^ln(5)-8^{1-y}-7x=9
(d^2y)/(dx^2)=(49(8^(2y-2)))/(3ln2)
y=sqrt((x+1)sqrt((x+2)sqrt((x+3)sqrt((x+4)
y'=[1/(2(x+1))+1/(4(x+2))+1/(8(x+3))+1/(16(x+1))]y
Find the equation of the tangent at (0, 1) of the curve,
y^3-xy^2+cos(xy)=2
y=1+1/3x