The first derivative of e to the x power is e to the power of x.
The anti-derivative of 1/x is ln|x| + C, where ln refers to logarithm of x to the base e and |x| refers to the absolute value of x, and C is a constant.
The derivative of ex is ex
That's because powers that involve the power "e", and logarithms to the base "e", are simpler than other powers or logarithms. For example: the derivative of ex is ex, while a derivative with other bases is more complicated; while the derivative of the natural logarithm (ln x, or logex) is 1/x.
e-x = 6Take the natural log of both sides:ln(e-x) = ln(6)-x = ln(6)x = -ln(6)So x = -ln(6), which is about -1.792.
y = e^ln x using the fact that e to the ln x is just x, and the derivative of x is 1: y = x y' = 1
e^[ln(x^2)]=x^2, so your question is really, "What is the derivative of x^2," to which the answer is 2x.
The derivative of e^u(x) with respect to x: [du/dx]*[e^u(x)]For a general exponential: b^x, can be rewritten as b^x = e^(x*ln(b))So derivative of b^x = derivative of e^u(x), where u(x) = x*ln(b).Derivative of x*ln(b) = ln(b). {remember b is just a constant, so ln(b) is a constant}So derivative of b^x = ln(b)*e^(x*ln(b))= ln(b) * b^x(from above)
The derivative of e^u(x) with respect to x: [du/dx]*[e^u(x)]For a general exponential: b^x, can be rewritten as b^x = e^(x*ln(b))So derivative of b^x = derivative of e^u(x), where u(x) = x*ln(b).Derivative of x*ln(b) = ln(b). {remember b is just a constant, so ln(b) is a constant}So derivative of b^x = ln(b)*e^(x*ln(b))= ln(b) * b^x(from above)
squared 3
y = (sinx)^(e^x) ln(y) = ln((sinx)^(e^x)) ln(y) = (e^x)ln(sinx) (1/y)dy = (e^x)(1/sinx)(cosx)+ln(sinx)(e^x)dx (1/y)dy = (e^x)(cotx)+ln(sinx)(e^x)dx dy = ((sinx)^(e^x))((cotx)(e^x)+ln(sinx)(e^x))dx dy = ((e^x)(sinx)^(e^x))(cotx+ln(sinx))dx
Yes, the function ln(x) where ln is the logarithm to base e.Yes, the function ln(x) where ln is the logarithm to base e.Yes, the function ln(x) where ln is the logarithm to base e.Yes, the function ln(x) where ln is the logarithm to base e.
The first derivative of e to the x power is e to the power of x.
It depends. If you mean (ln e)7, then the answer is 1, since (ln e) = 1. If you mean ln(e7), then the answer is 7, since ln(e7) = 7 (ln e) and (ln e) = 1.
Assuming that is the natural logarithm (logarithm to base e), the derivative of ln x is 1/x. For other bases, the derivative of logax = 1 / (x ln a), where ln a is the natural logarithm of a. Natural logarithms are based on the number e, which is approximately 2.718.
The definition of the natural log ln of a number is the power that you have to raise e to in order to get that number. Therefore, ln(2x+3) is the power you have to raise e to to get 2x + 3.
e^(3lnx)=e^[ln(x^3)]=x^3