2015年3月27日 · For f(x)=e^(2x^2), the derivative is f'(x)=4xe^(2x^2). To get this answer, we use the fact that the exponential function is its own derivative, together with the chain rule: For f(x)=e^(g(x)), the derivative is: f'(x)=e^(g(x)) g'(x), In differential operator notation: d/(dx)(e^u)=e^u (du)/(dx) For f(x)=e^(2x^2), the derivative is f'(x)=e^(2x^2)*(4x)=4xe^(2x^2).

2016年8月10日 · #rArre^x xxe^2=e^(x+2)# Answer link. Related questions. What is the Exponential Property Involving Products?

2016年8月25日 · ln e^(2x) = 2x As a Real valued function, x |-> e^x is one to one from (-oo, oo) onto (0, oo). As a result, for any y in (0, oo) there is a unique Real value ln y such that e^(ln y) = y. This is the definition of the Real natural logarithm. If t in (-oo, oo) then y = e^t in (0, oo) and from the above definition: e^(ln(e^t)) = e^t Since x |-> e^x …

2015年8月14日 · dy/dx= -2.e^(-2x) Understand e^((some expression)) is to be differentiated as (Differentiation of some expression). e^((some expression)) Differentiation of - 2x is = -2 Hence dy/dx= -2.e^(-2x)

2018年6月27日 · ln(2x+1)=2-ln(x) e^(ln(2x+1))=e^(2-ln(x)) 2x+1=e^2/x 2x^2+x=e^2 2x^2+x-e^2=0 x=(-1+-sqrt((1+8(e^2))))/4 2*(-1-sqrt((1+8(e^2))))/4+1<0 Therefore ln(2*(-1-sqrt((1+8(e^2 ...

Since the derivative of e^x is itself, the integral is simply e^x+c. The integral of other exponential functions can be found similarly by knowing the properties of the derivative of e^x.

2018年4月10日 · #int_0^1 e^(x^2)dx = 1/42 + 1/10 + 1/3 + 1 ~~1.457# A calculator should give an approximation of #1.463# , so our answer isn't too terrible. Increasing the number of terms of the maclaurin series in the application will make the approximation more precise.

2016年8月14日 · This can be done by substitution. See explanation. intxe^(x^2)dx=|(t=x^2),(dt=2xdx),(dt/2=xdx)|=int(e^t/2)dt=1/2int(e^tdt) =1/2e^t+C=1/2e^(x^2)+C

2018年5月23日 · Then du = 4dx and dx= (du)/4 I = 1/4 int (e^u + 1)/(e^u - 1)du Now use partial fractions. A/(e^u - 1) + B/1 = (e^u + 1)/(e^u - 1) A + Be^u - B = e^u + 1 We can readily see that B = 1 and therefore A = 2. Thus I = 1/4(int 2/(e^u - 1) + 1 du) I = 1/2ln|e^u - 1| + 1/4u + C I = 1/2ln|e^(4x) - 1| + x + C Hopefully this helps!

2015年5月25日 · We can rewrite the sum as: sum_(n=0)^oo n/(e(n^2)) = 1/e sum_(n=o)^oo n/n^2 = 1/e sum_(n=o)^oo 1/n Thus we can see that sum_(n=0)^oo 1/n is the Divergent Harmonic Series. Thus we have a scalar multiple of a Divergent series, thus we end up with a Divergent series. so: 1/e sum_(n=0)^oo 1/n is divergent Proof of the divergence of the Harmonic Series.