What is the derivative of 1÷x?
We know that, [math]1÷x=\frac{1}{x}=x^{-1}[/math]Derivative of [math]x^{n}=n(x)^{n-1}[/math][math]\therefore \frac{d}{dx}x^{-1}=-1(x)^{-1-1}[/math][math]\therefore \frac{d}{dx}x^{-1}=-1(x)^{-2}[/math][math]\therefore \frac{d}{dx}x^{-1}=-\frac{1}{x^2}[/math]Hence, we conclude that derivative of [math]1÷x[/math] is [math]-\frac{1}{x^2}[/math].Hope this helps! You can also get such questions solved and get the detailed solution within seconds using the Scholr app by just uploading a picture of the question and also get to be a part of an ever-growing community of students.
What is the derivative of 2^n?
ZERO n is usually reserved for a constant, possibly an integer. If n is a constant (any real number) then 2^n is also a constant and the derivative of a constant is 0. If x is a variable with domain of all real numbers and y = 2^n, then y' = 2^n(ln2) by basic derivative formulas.
What is the derivative of e^3?
If you consider e^x you get (1 + 1/n)^(nx) Expand this by the binomial theorem and you have (1 + 1/n)^(nx) = 1 + (nx)(1/n) + nx(nx-1)/2! (1/n)^2 + ... and carrying through the same process of putting the n's in the denominator into each term in the numerator, as described above, you obtain e^x = 1 + x + x^2/2! + x^3/3! + ... and differentiating this we get d(e^x)/dx = 0 + 1 + 2x/2! + 3x^2/3! + ... = 1 + x + x^2/2! + x^3/3! + ... = e^x --------------------------------------... e is an irrational number, thats why it isn't o try this Usually, the first thing about e that one learns in school is the definition e=lim n→∞ (1+1/n)^n. Later on one is taught if f(x)=e^x then f'(x)=e^x. Most school textbooks on mathematics however lack proof for this, even though it is quite simple. From the definintion of the derivative: f'(x)=(f(x+Δx)-f(x))/Δx when Δx→0 f'(x)=(e^(x+Δx)-e^x)/Δx f'(x)=(e^x*e^Δx-e^x)/Δx f'(x)=e^x*(e^Δx-1)/Δx Now, all that remain is to show that (e^Δx-1)/Δx→1 as Δx→0 To do this one inserts e=(1+1/n)^n when n→∞. f'(x)=e^x*((1+1/n)^n*Δx-1)/dx As n→∞ and Δx→0, we can get rid of one of the limits by writing n=1/Δx. This makes the above equation simplify to f'(x)=e^x*((1+Δx)^1 - 1)/Δx f'(x)=e^x*Δx/Δx=e^x just google "derivative of e^x
What is the derivative of 1/x?
1.) You can have memorized that it is [math]\frac{-1}{x^2}[/math]2.) you can rewrite it as [math]\frac{d}{dx}[x^{-1}][/math] and apply the power rule [math]\frac{d}{dx} [x^n] = nx^{n-1}[/math]so for our problem we would get [math]\frac{d}{dx}[x^{-1}]= -1 x^{-1-1} = -x^{-2} = \frac{-1}{x^2}[/math]3.) You can use the definition of the limit and evaluate [math]\lim_{h \to 0} \frac{\frac{1}{x+h}- \frac{1}{x}}{h} [/math]
What is the derivative of f(x)= 1 / x^n ?
if x = 1/x^n then if you bring the x^n up it becomes 1x^-n therefore taking the derivative of that will be 1(-n)x^(-n-1) *edited
How do you take the derivative of f(x)=(n+1)! ?
f(n)(0) = (n+1)! means that the nth derivative of f(x) with respect to x, evaluated at x=0, equals (n+1)! The Maclaurin series for f(x) is f(x) = f(0) + f'(0) x + f''(0) x²/2! + f'''(0) x³/3! + ... = ∞ Σ f(n)(0) x^n / n! n=0 You are told that f(n)(0) = (n+1)!, so putting that in the sum, ∞ Σ (n+1)! x^n / n! n=0 But this equals ∞ Σ (n+1) x^n n=0 So f(x) = 1 + 2x + 3x² + 4x³ + ... Use the Ratio Test to show that the radius of convergence is 1 (that is, the series converges for |x|<1)
What is the derivative of 1/(x+1) and x + 1/x ?
For the best answers, search on this site https://shorturl.im/tEsqI You have the derivatives wrong - I worked them below y = 0.5ln((1+x)/(1-x)) = 0.5ln[(1+x)(1-x)‾¹] Using the Chain Rule [ if f(x) = h(g(x)), then f'(x) = h'(g(x))g'(x) ] And d(lnx)/dx = 1/x y' = 0.5*(1/[(1+x)(1-x)‾¹])*d([(1+x)(1-x)‾¹])... y' = 0.5(1-x)(1+x)‾¹d([(1+x)(1-x)‾¹])/dx Using Product Rule d([(1+x)(1-x)‾¹])/dx = x(1-x)‾¹ + (1+x)(1-x)‾²(-1)(-1) = x(1-x)‾¹ + (1+x)(1-x)‾² So y' = 0.5(1-x)(1+x)‾¹[x(1-x)‾¹ + (1+x)(1-x)‾²] y' = 0.5x(1+x)‾¹ + 0.5(1-x)‾¹ = 0.5[x(1+x)‾¹ + (1-x)‾¹] y' = 0.5[(x(1-x) + (1+x))/(1+x)(1-x)] y' = 0.5[(x-x²+1+x)/(1-x²)] y' = -0.5(x²-2x-1)(1-x²)‾¹ Again using Product Rule y'' = -0.5[(2x-2)(1-x²)‾¹ + (x²-2x-1)(1-x²)‾²(-2x)(-1)] y'' = -0.5[2(x-1)(1-x²)‾¹ +2x(x²-2x-1)(1-x²)‾²] y'' = -(1-x²)‾²[(x-1)(1-x²) + x(x²-2x-1)] y'' = -(1-x²)‾²[x-x³-1+x²+x³-2x²-x] y'' = -(1-x²)‾²(-x²-1) y'' = (1+ x²)(1-x²)‾² And then y''' = 2x(1-x²)‾² + (1+ x²)(1-x²)‾³.(-2x)(-2) y''' = (1-x²)‾³[2x(1-x²) + 4x²(1+ x²)] y''' = 2x(1-x²)‾³[(1-x²) + 2x(1+ x²)] y''' = 2x(1-x²)‾³[1-x² + 2x+ 2x³] y''' = 2x(1-x²)‾³[2x³-x² + 2x+ 1] y''' = 2x(2x³-x² + 2x+ 1)(1-x²)‾³ So y = 0.5ln((1+x)/(1-x)) = 0.5ln[(1+x)(1-x)‾¹] y' = -0.5(x²-2x-1)(1-x²)‾¹ y'' = (1+ x²)(1-x²)‾² y''' = 2x(2x³-x² + 2x+ 1)(1-x²)‾³ The Taylor/Maclaurin series of (tanh)‾¹(x) (which is arctanh(x)) is arctanh(x) = Sum[(2n+1)‾¹.x^(2n+1)] And as MacLaurin series is defined as Sum((f^n(a)/n!)x^n), where f^n is the nth derivative of f(n) So for arctanh f^1 = x^3/3 - which is not the same as y' as son on for other derivatives
What is nth derivative of x^n/(x-1)?
I assume you mean[math]\frac{x^n}{x-1}[/math]If so, rewrite it as[math]x^{n-1}+x^{n-2}+\dots+x+1+\frac1{x-1}[/math]You can clearly see that all the terms but the last one will vanish after n derivatives.So what is the nth derivative of [math]\frac1{x-1}[/math]?The first is [math]\frac{-1}{(x-1)^2}[/math].The second is [math]\frac{2}{(x-1)^3}[/math].The third is [math]\frac{-6}{(x-1)^4}[/math].By now, you can spot the pattern and see the nth derivative will be [math]\frac{(-1)^nn!}{(x-1)^{n+1}}[/math].Just like the original function, it is only defined when[math] x\neq 1[/math].
What is the nth derivative of cosx?
Let [math]f = \cos x[/math]Let [math]f^{n}[/math] be the nth derivative of [math]f[/math].[math]f^{1} = - \sin x[/math][math]f^{2} = - \cos x[/math][math]f^{3} = \sin x[/math][math]f^{4} = \cos x[/math][math]f^{5} = - \sin x[/math][math].[/math][math].[/math][math].[/math][math]f^{n} = - \cos x[/math], if [math]n[/math] is even and not a multiple of [math]4[/math][math]f^{n} = \cos x[/math], if [math]n[/math] is even and is a multiple of [math]4[/math][math]f^{n} = \sin x[/math], if [math]n[/math] is odd and is a multiple of [math]3[/math][math]f^{n} = - \sin x[/math], if [math]n[/math] is odd and not a multiple of [math]3[/math]
What is the nth derivative of x^2n?
[math]\displaystyle\qquad\frac{\mathrm d}{\mathrm dx}x^{2n}=2nx^{2n-1}[/math][math]\displaystyle\qquad\frac{\mathrm d^2}{(\mathrm dx)^2}x^{2n}=2n(2n-1)x^{2n-2}[/math][math]\displaystyle\qquad\frac{\mathrm d^3}{(\mathrm dx)^3}x^{2n}=2n(2n-1)(2n-2)x^{2n-3}[/math][math]\displaystyle\qquad\frac{\mathrm d^n}{(\mathrm dx)^n}x^{2n}=\prod_{k=0}^{n-1}(2n-k)*x^{2n-n}[/math]We can represent that product by the division of two factorials. [math]\displaystyle a(a-1)(a-2)\cdots(a-b)=\frac{a(a-1)(a-2)\cdots(3)(2)}{(a-b-1)(a-b-2)\cdots(3)(2)}=\frac{a!}{(a-b-1)!}[/math][math]\displaystyle\qquad\frac{\mathrm d^n}{(\mathrm dx)^n}x^{2n}=\frac{(2n)!}{(2n-(n-1)-1)!}x^n[/math][math]\displaystyle\qquad\frac{\mathrm d^n}{(\mathrm dx)^n}x^{2n}=\frac{(2n)!}{n!}x^n[/math]