The antiderivative of 9sinx is simply just -9cosx. It is negetive because the derivative of cosx should have been -sinx, however, the derivative provided is positive. Therefore, it means that there should be a negative with cosx in order to make that sinx postive. (negative times negative eguals positive)
take out the constant -2 then take the intergral of cosx this will give you sinx your answer is -2sinx
The anti derivative of negative sine is cosine.
When you solve for the 2nd derivative, you are determining whether the function is concave up/down. If you calculated that the 2nd derivative is negative, the function is concave down, which means you have a relative/absolute maximum, given that the 1st derivative equals 0. To understand why this is, think about the definition of the 2nd derivative. It is a measure of the rate of change of the gradient. At a maximum, the gradient starts positive, becomes 0 at the maximum itself and then becomes negative, so it is decreasing. If the gradient is going down, then its rate of change, the 2nd derivative, must be negative.
- e^- X
d/dx(-cosx)=--sinx=sinx
The antiderivative of 9sinx is simply just -9cosx. It is negetive because the derivative of cosx should have been -sinx, however, the derivative provided is positive. Therefore, it means that there should be a negative with cosx in order to make that sinx postive. (negative times negative eguals positive)
-sinx
Trig functions have their own special derivatives that you will have to memorize. For instance: the derivative of sinx is cosx. The derivative of cosx is -sinx The derivative of tanx is sec2x The derivative of cscx is -cscxcotx The derivative of secx is secxtanx The derivative of cotx is -csc2x
d/dx(sinx-cosx)=cosx--sinx=cosx+sinx
2
derivative (7cosx) = -ln(7) 7cosx sinx dx
The derivative of sin(x) is cos(x).
Using the Chain Rule :derivative of (sinx)2 = 2(sinx)1 * (derivative of sinx)d/dx (Sinx)2 = 2(sinx)1 * [d/dx (Sinx)]d/dx (Sinx)2 = 2(sinx) * (cosx)d/dx (Sinx)2 = 2 (sinx) * (cosx)d/dx (Sinx)2 = 2 sin(x) * cos(x)
(cosx)^2-(sinx)^2
y=1/sinxy'=(sinx*d/dx(1)-1*d/dx(sinx))/(sin2x)y'=(sinx*0-1(cosx))/(sin2x)y'=(-cosx)/(sin2x)y'=-(cosx/sinx)*(1/sinx)y'=-cotx*cscx
f(x)=sinx+cosx take the derivative f'(x)=cosx-sinx critical number when x=pi/4