f(x)=(x+1)^2 find the slope of the tangent line at x=2
f'(x)=2(x+1), 2(2+1)=6
Lim x->3 of (3x^2)
27
dy/dx(3x^2+y)
=6x+dy/dx
int(3x^2)dx
x^3+C
avg value of int from 0-2 of (siny)dy
1/2[-cos(2)+1]
f(x)=sin(3x)
f'(x)= 3cos(3x)
lim x->5 of ([2x^3]/x)
50
dy/dx{(3x^2)*(y^2)}
=(6xy^2)+(3x^2*2ydy/dx)
int(sin(x))dx
-cos(x)+C
int(3cos2x)dx
(3/2)sin(2x)+C
f(x)=4cos(3x/2)
f'(x)=-4sin(3x/2) (3/2)=
-6sin(3x/2)
lim x->2 of ([7x^3]/9x^2)
56/36
dy/dx{π₯^3β’*π¦^5 +3β’π₯ +10}
{(3x^2*y^5)(5y^4 dy/dx*x^3)+3}
int{5x^2+sin(3x)}dx
(5x^3)/3 -1/3cos(3x)+C
int([-1/2][cos4x])dx
-1/8sin4x+C
πβ‘(π₯) =(6β’π₯^2+7β’π₯)^4
πβ²β‘(π₯)=4β’(6β’π₯^2+7β’π₯)^3β’(12β’π₯+7)=
4β’(12β’π₯+7)β’(6β’π₯^2+7β’π₯)^3
lim x->pi of sec(X)
-1
dy/dx{(x^4)/(y^2)}
{[(4x^3*y^2)-(x^4*2y*dy/dx)]/(y^2)^2 }
1/3 int{(7y^4)}dy
1/15(7y^5)+C
int(tanx)dx
-ln(abs val[cosx])+C
ββ‘(π§) =sinβ‘(π§^6) +sin^6β‘(π§)
ββ²β‘(π§)=6β’π§^5β’ cosβ‘(π§^6)+ 6β’sin^5β‘(π§)β’cosβ‘(π§)
lim x->4pi/3 of tan^2(x)
3
dy/dx{(siny)/(x^3)}
{[(cosy*dy/dx*x^3)-(3x^2*siny)]/(x^3)^2}
int{(x+x^1/3)(4-x^2)}dx
=int{4x-x^3+4x^1/3-x^7/3}
=2x^2 -1/4(x^4) +3x^4/3 -3/10(x^10/3)+C
int[du/(a^2+u^2)]
1/a tan^-1(u/a)+C