f'(x)xy=yx(y-1)
f'(x)2=2x1=2x
derivative of sec2(x)=2tan(x)sec2(x)
the derivative of 3x is 3 the derivative of x cubed is 3 times x squared
2xsin2x+2x2sinxcosx
f(x)=(pi2)x=pi2x. The derivative of kx=ln(k)*kx, so f'(x)=2ln(pi)*pi2x (with chain rule).
f(x) = 3x2 + 5x + 2fprime(x) = 6x + 5
derivative of sec2(x)=2tan(x)sec2(x)
the derivative of 3x is 3 the derivative of x cubed is 3 times x squared
x squared
10 x
Let k = 0 9x18 squared x 17 x 18 k is a constant. Its anti-derivative is kx + C, where C is a constant. The anti-derivative squared is (kx+ C) squared.
f(x) = (1 - x^(2) ^(1/2) Let y= (1 - x^(2)) ^(1/2) Use Chain Rule dy/dx = dy/du X du/dx Let u = 1 - x^(2) Hence y = u^(1/2) dy/du = (1/2)u^(-1/2) du/dx = -2x Hence dy/dx = dy/du X du/dx = (1/2)u^(-1/2) X ( -2x) dy/dx = (1/2)u( 1 -x^(2)^(-1/2) X ( -2x) Tidying up dy/dx = (-2x/(2(1 - x^(2))^(1/2)) dy/dx = -x/ [(1 - x^(2)]^(1/2)
2xsin2x+2x2sinxcosx
-(1/2)X^2 [negative half X squared]
e^[ln(x^2)]=x^2, so your question is really, "What is the derivative of x^2," to which the answer is 2x.
It is -1 over x-squared.
14x
negative cotangent -- dcot(x)/dx=-csc^2(x)