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exponential function derivative : The derivative of the exponential function is equal to the value of the function . ..... More from Wikipedia
Derivative of Exponential Function
If f(x) = e x , then ..
If f(x) = e x , then ..Derivative of a Function
Derivative of a Function - So far we have discussed the derivative of a function f(x) at a point 'a' which is in the domain of f. Suppose we want to find the derivative of the same function at a different point 'b', then we have to compute..
Derivatives of Exponential Functions - I give the basic formulas and do a few examples involving derivatives of exponential functions. For more free math videos, visit JustMathTutoring.com
In this video, I give the formulas for finding derivatives of logarithmic functions and use them to find derivatives! For more free math videos, visit JustMathTutoring.com
Question : From the topic Derivative of exponential function.
For the curve y=e^ -4x find the tangent line parallel to the line 2x+y=7.
Answer : First, find the derivative of e^(-4x). This will give you the slope of the curve: y = e^-4x y' = e^(-4x) * -4 = -4 * e^(-4x) Now, find the slope of 2x + y = 7 y = -2x + 7 your slope is -2 Now, find when -4 * e^(-4x) = -2 -4 * e^(-4x) = -2 e^(-4x) = 1/2 -4x = ln(1/2) x = -ln(1/2) / 4 x = 0.17328679513998632735430803036454 (I'll keep it in it's exact form for the time being). So, when x = -ln(1/2) / 4, the slope of e^(-4x) is identical to the slope of 2x + y = 7 So, now we need a line that touches the curve of e^(-4x) when x = -ln(1/2)/4 and it has a slope of -2. So, first we need to know what e^(-4x) equals when x = -ln(1/2) / 4 e^(-4 * -ln(1/2)) = e^(ln(1/2)) = 1/2 Now, we plug in our values for x and y and solve for the y-intercept (b) y = mx + b m = -2 y = 1/2 x = -ln(1/2) / 4 1/2 = -2 * -ln(1/2) / 4 + b 1/2 = ln(1/2) / 2 + b (1 - ln(1/2)) / 2 = b So the tangent line that is parallel to 2x + y = 7 is this: y = -2 * x + .... More from Yahoo Answers
Answer : First, find the derivative of e^(-4x). This will give you the slope of the curve: y = e^-4x y' = e^(-4x) * -4 = -4 * e^(-4x) Now, find the slope of 2x + y = 7 y = -2x + 7 your slope is -2 Now, find when -4 * e^(-4x) = -2 -4 * e^(-4x) = -2 e^(-4x) = 1/2 -4x = ln(1/2) x = -ln(1/2) / 4 x = 0.17328679513998632735430803036454 (I'll keep it in it's exact form for the time being). So, when x = -ln(1/2) / 4, the slope of e^(-4x) is identical to the slope of 2x + y = 7 So, now we need a line that touches the curve of e^(-4x) when x = -ln(1/2)/4 and it has a slope of -2. So, first we need to know what e^(-4x) equals when x = -ln(1/2) / 4 e^(-4 * -ln(1/2)) = e^(ln(1/2)) = 1/2 Now, we plug in our values for x and y and solve for the y-intercept (b) y = mx + b m = -2 y = 1/2 x = -ln(1/2) / 4 1/2 = -2 * -ln(1/2) / 4 + b 1/2 = ln(1/2) / 2 + b (1 - ln(1/2)) / 2 = b So the tangent line that is parallel to 2x + y = 7 is this: y = -2 * x + .... More from Yahoo Answers
Question : Use the Chain Rule in conjunction with the exponential or Logarithm Derivative Rule to find the derivative for each of the following:
f)y=log_10(1+x/1-x)
g)f(x)=7^(x)^(2)
h)v=log_2( t^(2)+3t)
i)y=3^(x)^((2)+3)
Answer : g) f(x)=7^(x^2) Let u=x^2 du/dx = 2x f(x)=7^u f'(x) = 7^u ln(7) du/dx f'(x)= 7^(x^2) ln(7) (2x) =2ln(7) x 7^(x^2) Note: The derivative of a^x is a^x ln(a) where a is any constant. h) v= log-2 sqrt(t^2+3t) v= ln sqrt(t^2+3t) / ln(2) v= (1/ln(2)) ln (t^2+3t)^(1/2) v = [1/ln(2)] (1/2) ln (t^2+3t) let u = t^2+3t du/dt = 2t+3 v = [1/ln(2)] (1/2) ln(u) dv/dt = 1/ln(2)] (1/2) (1/u) du/dt dv/dt = [1/ln(2)] (1/2) [2t+3) / t^2+3t] i) y=3^(x^2+3) let u=x^2+3 du/dx =2x y=3^u dy/dx = 3^u ln(3) du/dx dy/dx = 2 ln(3) x 3^(x^2+3).. More from Yahoo Answers
Answer : g) f(x)=7^(x^2) Let u=x^2 du/dx = 2x f(x)=7^u f'(x) = 7^u ln(7) du/dx f'(x)= 7^(x^2) ln(7) (2x) =2ln(7) x 7^(x^2) Note: The derivative of a^x is a^x ln(a) where a is any constant. h) v= log-2 sqrt(t^2+3t) v= ln sqrt(t^2+3t) / ln(2) v= (1/ln(2)) ln (t^2+3t)^(1/2) v = [1/ln(2)] (1/2) ln (t^2+3t) let u = t^2+3t du/dt = 2t+3 v = [1/ln(2)] (1/2) ln(u) dv/dt = 1/ln(2)] (1/2) (1/u) du/dt dv/dt = [1/ln(2)] (1/2) [2t+3) / t^2+3t] i) y=3^(x^2+3) let u=x^2+3 du/dx =2x y=3^u dy/dx = 3^u ln(3) du/dx dy/dx = 2 ln(3) x 3^(x^2+3).. More from Yahoo Answers
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