The 50th derivative of a function?
Well, there are three main ways that I know of to find the 50th derivative of a function: 1) Actually take the derivative 50 times. On a test, obviously this one is not what we're going to do. 2) Take the derivative a few times and look for a pattern. For example, we can find the derivative of f(x) = e^(3x) this way: f(x) = e^(3x) f ' (x) = 3e^(3x) f '' (x) = 9e^(3x) f ''' (x) = 27e^(3x) f^(4) (x) = 81e^(3x) It looks like the pattern is that, each time we take the derivative, we multiply the function we had before by 3. Multiplying f(x) by 3 fifty times gives (3^50)e^(3x). So the 50th derivative of e^(3x) is (3^50)e^(3x). As another sort of pattern, if you have a polynomial of degree 49 or less, then its 50th derivative is 0. So, for example, the 50th derivative of x^49 + 800x^17 + 98435462x^7 is 0. 3) Use the Taylor series. (If you don't know what a Taylor series is, then ignore this part.) Recall that, if we have a Taylor series centered at a for a function f(x), then in that series, the coefficient of (x - a)^n is equal to f^(n)(a) / n!, where f^(n)(a) means the nth derivative of f, evaluated at the value a. So, for example, we can find the 50th derivative of f(x) = cos (2x), evaluated at 0, by first finding the Taylor series for cos (2x) centered at a = 0. To do this, first recall that the Taylor series for cos (x) centered at a = 0 is cos (x) = Σ (-1)^n x^(2n) / (2n)! Thus, to find the Taylor series for cos (2x), we may plug in 2x where x was before: cos (2x) = Σ (-1)^n (2x)^(2n) / (2n)! Let us simplify the right side a bit: cos (2x) = Σ (-1)^n * 2^(2n) * x^(2n) / (2n)! To find the 50th derivative of cos (2x) evaluated at 0, we can look at the coefficient of x^50. To find the coefficient of x^50 in this series, notice that in the n = 25 term, we have (-1)^25 * 2^(50) * x^(50) / (50)!. We know, then, from the general form of the coefficient of x^50 in a Taylor series, that f^(50)(0) / (50!) = (-1)^25 * 2^(50) / 50!. Solving this equation for f^(50)(0) gives -2^50. So the 50th derivative of cos (2x), evaluated at 0, is -2^50.
Find the partial derivatives of the function f(x,y)=xye^(−2y)?
To take partial derivatives, you treat the other variable as a constant. For example, fx(x,y) is taking the derivative with respect to x, while treating y as constant. Then once you have fx(x,y), fxy(x,y) would be taking the derivative of fx(x,y) with respect to y. Both fxy(x,y) and fyx(x,y) should yield the same answer so you can tell if any errors occur. Basically, this says it does not matter in which order you take derivatives, it is the same end result. fx(x,y)=ye^(-2y) fy(x,y)=xe^(-2y)-2xye^(-2y) fxy(x,y)=e^(-2y)-2ye^(-2y) fyx(x,y)=e^(-2y)-2ye^(-2y) Same solutions in the end, so we know it's correct.
What is the derivative of [math]e^{2x}[/math]?
[math]y=e^{2x}[/math]Since we have that 2 there, it can't just be [math]e^{2x}[/math]When working with the chain rule [math]f(g(x)) = f'(g(x))g'(x)[/math]Basically you take the derivative of the outside, multiplied by the inside.[math]\frac{\mathrm{d}}{\mathrm{d} x} sin(u) = u'cos(u)[/math]Let's say that [math]u = 2x[/math][math]\frac{\mathrm{d} }{\mathrm{d} x}e^{u}=u' * e^{u}[/math]Since u = 2x, u' = 2 [math]\frac{\mathrm{d} }{\mathrm{d} x}e^{2x} = 2e^{2x}[/math]But let's try another way. Since [math]y = e^{2x}[/math]Then [math]ln(y) = 2x[/math]Now, take the derivatives of both sides. [math]\frac{\mathrm{d} }{\mathrm{d} x}ln(y) = 2[/math][math]ln(u) = \frac{u'}{u}[/math][math]y=u[/math] then [math]\frac{\mathrm{d} }{\mathrm{d} x}ln(y) = \frac{y'}{y}[/math]Multiply by y.[math]y' = y*2[/math][math]y=e^{2x}[/math]Substituting in, [math]y' = 2 * e^{2x}[/math]
Are there any functions whose derivative is equal to itself?
Any function of the form [math]y=ae^x[/math] should work fine.Statement: Any function of the form [math]y=ae^x[/math] has the same derivative as itselfProof:Let [math]\dfrac{dy}{dx}=y[/math], where [math]y=f(x)[/math] is an arbitrary function[math]\implies \displaystyle \int \dfrac{dy}{y}=\int dx[/math][math]\implies \ln y=x+C[/math][math]\implies \ln y=x+\ln a[/math], where [math]C=\ln a[/math][math]\implies \ln y-\ln a=x[/math][math]\implies \ln \left(\dfrac{y}{a}\right)=x[/math][math]\implies \dfrac{y}{a}=e^x[/math][math]\implies y=ae^x[/math]Conclusion: [math]y=ae^x[/math] is the only type of function whose derivative is itself.Thanks for the A2A
Find an equation of the tangent line to the graph of f(x) = 1/ x-1 at the point (2,1).?
If f(x) = 1/x -1 then we find the derivative of this function f'(x) = -1/x^2 is that derivative which is the slope of our original function f(x) Since our point is (2,1) we use x=2 to find the slope f'(2) = -1/4 We next use the point-slope formula to find the equation of the tangent y-y0 = m(x-x0) y - 1 = -1/4(x - 2) y = -1/4x + 2 + 1 y = -1/4x + 3 This doesn't match any choice A-E which might mean that I got the function wrong. Let me try again, f(x) = 1/(x-1) we find the derivative of this function f'(x) = -1/(x-1)^2 and at point (2,1) we have f'(x) = -1 then y - 1 = -1(x-2) y - 1 = -x +2 y = -x + 3 y + x - 3 = 0 I am going to guess that this is the correct function f(x). Using a couple of braces would prevent confusion. In any case, D. none of the above is the answer
F(x) = -10x. Find the derivative using the limit process.?
All you need is this equation: f '(x) = lim h->0 (f(x + h) - f(x)) / h lim (-10(x + h) - -10x) / h h->0 lim (-10x - 10h + 10x) / h h->0 lim (-10h) / h h->0 -10
Consider the equation. f(x) = e^(2x) + e^(−x)?
(a) Find the intervals on which f is increasing. (Enter your answer using interval notation.) Find the interval on which f is decreasing. (Enter your answer using interval notation.) (b) Find the local minimum value of f. (c) Find the interval on which f is concave up. (Enter your answer using interval notation.)
What is the derivative of modulus function?
Whenever you see a modulus function then the best thing to do is to break it in parts if function has a negative value for any value of x , else just ignore the modulus operation and proceed. Lets understand it with examples.Let our function be f(x) ( where f(x) < 0 has solutions ) , its modulus is |f(x)|then for its derivativewe break it inf(x)>0Its derivative is+ d f(x) /dxf (x) < 0Its derivative is- d f(x) /dxf(x) = 0The derivative doesn’t exist since Left Hand derivative ( f(x) < 0) is not equals Right hand derivative (f(x) > 0). We can see that the sign of derivative depends on the sign of the function.So,To shorten the procedure done and avoid all the above fuss ( ONLY FOR FUNCTIONS WHICH HAVE NEGATIVE VALUES FOR ANY x )we write the derivative of f(x) as(d f(x) / dx ) * f(x) / | f(x) |this way we keep the signs of all cases and remove all x from the domain of derivative for which f(x) =0.EXAMPLEderivative of |sin x| is(cos x ) * (sin x) / |sin x|BUT for derivative of |x^2| we do not use this method we just ignore modulus operation and write its derivative as 2x.
Let f be the function defined by f(x) =x^3 + x. If g(x) is the inverse of f(x) and g(2) =1, what is the value of the derivative of g at x=2?
Let’s do this somewhat systematically.We should define things in Leibniz notation first, for mathematical convenience:Let [math]y_1 = f(x) = x^3+x[/math], and let [math]y_2 = g(x) = f^{-1}(x)[/math]To find the inverse of a function, all we need to do is to reflect the function about the line [math]y=x[/math], which basically maps all points [math](x,y)[/math] to [math](y,x)[/math]. In effect, we are swapping the roles of [math]x[/math] and [math]y[/math] as independent and dependent variables. We can apply this to [math]y_1 = x^3+x[/math] to get:[math]x = {y_2}^3+y_2[/math]Hey, we have some information about [math]g’(x)[/math], so maybe it will help differentiating both sides of that equation (make sure to do the chain rule correctly!):[math]\frac{d}{dx}x = \frac{d}{dx}({y_2}^3+y_2)[/math][math]1 = 3{y_2}^2\frac{dy_2}{dx} + \frac{dy_2}{dx}[/math][math]1 = (3{y_2}^2+1)\frac{dy_2}{dx}[/math][math]\frac{dy_2}{dx} = \frac{1}{3{y_2}^2+1}[/math]This sort of manipulation is called implicit differentiation, and it really helps us here when we want the derivative of a function when we can’t (or is difficult to) solve a function in closed form.Now - funnily enough - we are almost done! It is now a matter of plugging and chugging. We are looking for [math]g’(2) = \frac{dy_2}{dx}(2)[/math], given that [math]g(2) = 1[/math]. So let’s revert back to the notation introduced in the problem:[math]g’(x)= \frac{1}{3{(g(x))}^2+1}[/math][math]g’(2)= \frac{1}{3{(g(2))}^2+1}[/math][math]g’(2)= \frac{1}{3{(1)}^2+1}[/math][math]g’(2)= \frac{1}{4}[/math]
What’s the nth derivative of f(x) =1/x? Is there a formula?
To find the [math]n[/math]th derivative, we must look for the first four derivatives and see how they change.[math]\begin{align}f(x) = \dfrac{1}{x}\end{align}\tag*{}[/math][math]f'(x) = -x^{-2}[/math][math]f''(x) = 2x^{-3}[/math][math]f'''(x) = -2 \cdot 3x^{-4}[/math][math]f''''(x) = 2 \cdot 3 \cdot 4x^{-5}[/math]Now, we look for a pattern. First of all, we can see that there are alternating signs. So, to show this we must have a [math](-1)^n[/math]. After that, notice that we have the increasing products and the integer number being multiplied to that. This represents a factorial, [math]n![/math]. Finally, we see that we have decreasing negative exponents even though we plug in a positive integer. So, if we think about it we must start with a [math]-n[/math] and subtract one. So, if we manipulate it a little bit, we get [math]-[/math][math](n + 1)[/math]. Putting all this stuff together, we have…[math]\begin{align}f^n(x) = (-1)^n (n!)x^{-(n + 1)}\end{align}\tag*{}[/math]