2016年3月23日 · In order to simplify this, we need to use the difference law: loga − logb ≡ log(a b) ln(−e) −ln(− 1 e) = ln(−e − 1 e) = ln(e2) = 2lne = 2 We could also use the exponent and …

2015年4月26日 · In fact, you can rewrite that function as follows: c (x)=ln (1+e^x/1-e^x)=ln (1+e^x-e^x)=ln (1)=0 The derivative of a constant function is 0, so c' (x)=0 I'll try to infer the function …

2015年12月26日 · The derivative is d ln(1 + e2x) dx = (1 + e2x)' 1 +e2x = 2 ⋅ e2x 1 +e2x Finally dy dx = 2 ⋅ e2x 1 +e2x

2016年5月21日 · 1/5 We can rewrite the argument of the natural logarithm function: lim_ (xrarroo)ln (1+e^x)/ (5x)=lim_ (xrarroo)ln (e^x (e^-x+1))/ (5x) Use the logarithm rule ln (ab ...

2015年4月28日 · You can use the Chain Rule to derive ln first (as it is) and then multiply by the derivative of the argument as: y' = 1 1 + e2x ⋅ 2e2x = 2e2x 1 +e2x

2018年1月19日 · ln (e^x-1)-x We can write the argument as a fraction. We first need to get rid of the negative exponent. ln (1−e^−x) = ln (1-1/e^x) = ln ( (e^x-1)/e^x) From here, you use the …

2015年9月27日 · What is the derivative of ln( ex 1 + ex)? Calculus Differentiating Logarithmic Functions Differentiating Logarithmic Functions with Base e 2 Answers

2016年7月29日 · Calculus Differentiating Logarithmic Functions Differentiating Logarithmic Functions with Base e

2016年12月30日 · ln and e are inverse functions. Each operator is the inverse of the other. ln (e^a)=a. Here, ln(1 √e) = ln(e− 1 2) = − 1 2.

2016年1月25日 · How do you differentiate f (x) = 1 (ln(1 − (e−cos(x2))))3 2 using the chain rule?