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 might become easier if you split the problem in half. Since we know that d/dx[f(x)-g(x)] is the same as the derivative of d/dx[f(x)] - d/dx[g(x)], you can rewrite your problem as d/dx(e)- d/dx(2x). (it's common usage to write coefficients in from of variables, but that's just notation). The derivative of e is simply 0 (since e is a constant) and the derivative of 2x is 2. so d/dx(e-2x)=0-2=-2
-2
d/dx sec(2x) = 2sec(2x)tan(2x)
-2
e^(-2x) * -2 The derivative of e^F(x) is e^F(x) times the derivative of F(x)
e^[ln(x^2)]=x^2, so your question is really, "What is the derivative of x^2," to which the answer is 2x.
e^[ln(x^2)]=x^2, so your question is really, "What is the derivative of x^2," to which the answer is 2x.
You are supposed to use the chain rule for this. First step: derivative of root of sin2x is (1 / (2 root of sin 2x)) times the derivative of sin 2x. Second step: derivative of sin 2x is cos 2x times the derivative of 2x. Third step: derivative of 2x is 2. Finally, you need to multiply all the parts together.
It might become easier if you split the problem in half. Since we know that d/dx[f(x)-g(x)] is the same as the derivative of d/dx[f(x)] - d/dx[g(x)], you can rewrite your problem as d/dx(e)- d/dx(2x). (it's common usage to write coefficients in from of variables, but that's just notation). The derivative of e is simply 0 (since e is a constant) and the derivative of 2x is 2. so d/dx(e-2x)=0-2=-2
-2
I'm not sure what you're asking. The derivative of sin(2x^2) is 4xcos(2x^x)dx.The derivative of (sin(2x^2)^2) is 8xsin(2x^2)cos(2x^2)dx.
It is -2*exp(-2x)
2x
-2
3
d/dx sec(2x) = 2sec(2x)tan(2x)