d/dx(sinx-cosx)=cosx--sinx=cosx+sinx
(cosx)^2-(sinx)^2
d/dx(-cosx)=--sinx=sinx
d/dx (-cscx-sinx)=cscxcotx-cosx
It is cos2x that is, "cos-squared x".
The derivative of 2^x is 2^x * ln2 so the derivative of 2^cosx * ln2 multiplied by d/dx of cox, which is -sinx so the derivative of the inside function is -sinx * 2^cosx *ln2. As to the final question, using the chain rule, d/dx (2^cosx)^0.5 will equal half of (2^cosx)^-0.5 * -sinx * 2^cosx * ln2
-sinx
Take the derivative term by term. d/dx(X - cosX) = sin(X) ======
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
The derivative is 1/(1 + cosx)
The derivative of cos(x) is negative sin(x). Also, the derivative of sin(x) is cos(x).
(cosx)^2-(sinx)^2
negative sin(x)
d/dx cosx=-sin x
f(x)=sinx+cosx take the derivative f'(x)=cosx-sinx critical number when x=pi/4
f(x) sinx f`(x) = cosx