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second derivative of x raise to e raise x

second derivative

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Find Derivative of y = x^x. Classic - analyzemath.com


Find Derivative of y = x^x : Find Derivative of y = x x. A calculus tutorial on how to find the first derivative of y = x x for x > 0. Note that the function defined ...
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The derivative of x to the x - Math Central


I have another problem with regards to the derivative, what about the derivative of x to the power x? Cher . Hi Cher,
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Derivative of 2ˣ (old) (video) | Khan Academy


Derivative of 2ˣ (old) ... , hold on a second. We know how to take the derivative of e to the x. ... So if we actually raise e to that power, ...
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Derivative of e^x | Wyzant Resources


This function is unusual because it is the exact same as its derivative. This means that for every x value, the ... Proof of e x by Chain Rule and Derivative of the ...
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calculus - derivative of x^x^x... to infinity? - Mathematics ...


... in my attempt to calculate it's derivative, I did the following: $$y=x ... This is the second ... before actually deriving it. As a side remark, looking at $x=e^ ...
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Derivative Calculator: Solve Derivatives with Wolfram|Alpha


Online Derivative Calculator Solve derivatives with Wolfram ... Alpha is a great calculator for first, second, and third derivatives; derivatives at a point; ...
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What is the second derivative of $e^{(x^y)}$ - Mathematics ...


What is the second derivative of $e^{(x^y)}$ ... Raise Your Voice: Proclaim the Need for Net Neutrality. Related. 1. On the meaning of the second derivative. 2.
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What is the second derivative of f(x)= ln (x^3+e^x)? | Socratic


What is the second derivative of #f(x)= ln (x^3+e^x)#? ... Calculate the heat required to raise the ... Answer 19 minutes ago. What is the axis of symmetry and ...
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second derivative of x raise to e raise x


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Find the second derivative for 5x^3+7x^2+6x

5x^3 + 7x^2 + 6x To differentiate you take away one from the power and then multiply by it.   So if you differentiate y = x^n you get: dy/dx = (n-1)x^(n-1)   And the second derivative is simply the derivative of dy/dx : So if you differentiate dy/dx you get: d^2y/dx^2 = (n - 2)(n - 1)x^(n - 2)   So with this information you should be able to answer your question (Hint: 6x differentiated is 6 , and 6 differentiated is 0 )
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to find nth derivative of tan ^ ( - 1 ) ( x / a ) with solution

Let y=tan^-1(x/a), then tan y = x/a. sec^2(y)dy/dx=1/a so dy/dx=1/(a(sec^2(y))=1/(a(1+tan^2(y))=1/(a(1+(x^2/a^2))=1/(a+x^2/a)=(a+x^2/a)^-1 Write y'=dy/dx for convenience, so y'' is the second derivative and y'=(a+x^2/a)^-1 or sec^2(y)y'=1/a. Let u=v=sec y and w=y', then sec^2(y)dydx=1/a can be expressed uvw=1/a. Therefore we can get the second derivative by differentiating further: d/dx(u.v.w.)=v.w.du/dx+u.w.dv/dx+u.v.dw/dx=0 (differentiation by parts) 2sec(y).y'.sec(y)tan(y)y'+sec^2(y).y''=0 2sec^2(y)tan(y)y'^2+sec^2(y)y''=0, which we can divide through by sec^2(y): tan(y)y'^2+y''=0 making y''=-tan(y)y'^2=-x/a((a+x^2/a)^-2) So far, there's no obvious pattern in the derivatives which would suggest the nth derivative. Let's continue by finding the 3rd derivative, using the derivatives of y rather than converting to expressions in x. y'''=d/dx(-tan(y)y'^2) To calculate d/dx(y'^2) we can differentiate by parts and the result is 2y'y''. If we differentiate the right side of this equation we get y'''=-sec^2(y)y'-2tan(y)y'y''=-(1+tan^2(y))y'-2tan(y)y'y''=-y'(1+tan^2(y)+2tan(y)y''). Or, since we know that sec^2(y)y'=1/a, we can substitute to simplify the third derivative: y'''=-1/a-2tan(y)y'y''. Also, y''=-tan(y)y'^2 and tan(y)=x/a, so a simpler expression for y'''=-1/a-(2x/a)y'y''. More to follow...
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given: y= f(x) = x^3 - 3x +2 determine dy/dx, the second derivative, turning points, min and max, point of inflection

f'(x)=dy/dx=3x^2-3; when 3x^2-3=0 there is a turning-point (the gradient is horizontal). So x^2=1 and x=1 and -1. Second derivative f"(x)=6x. At x=1 f"(1)>0 (minimum) and at x=-1 f"(-1)<0 (maximum). When x=1 y=1-3+2=0; when x=-1 y=-1+3+2=4. The coords for min are (1,0) and max (-1,4).
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if sinx+cosy=1 then how will i solve its second order derivative?

sin x + cos y =1, so cos y = 1-sin x, so sin y = sqrt(1-(1-sin x)^2)=sqrt(2sin x - sin^2x) If we differentiate with respect to x we get cos x - sin y.dy/dx=0. So we can write dy/dx=(cos x)/(sin y). Differentiate again and we get -sin x - cosy(dy/dx)(dy/dx) - sin y.d2y/dx2=0 [d2y/dx2 represents the second derivative] -sin x - (1-sin x)cos^2x/sin^2y - sin y.d2y/dx2=0 when we make some substitutions. From this we get the second derivative: sin y.d2y/dx2=-(sin x + (1-sin x)cos^2x/(2sin x-sin^2x)=-(2sin^2x-sin^3x + (1-sin x)cos^2x)/(2sin x-sin^2x) The final substitution gives d2y/dx2=-(2sin^2x-sin^3x + (1-sin x)cos^2x)/(2sin x-sin^2x).sqrt(2sin x - sin^2x) =-(2sin^2x-sin^3x+(1-sin x)(1-sin^2x))/(sin x(2-sin x))^(3/2) =-(2sin^2x-sin^3x+1-sin^2x-sin x+sin^3x)/(sin x(2-sin x))^(3/2) =-(sin^2x-sin x+1)/(sin x(2-sin x))^(3/2) or (sin x-sin^2x-1)/(sin x(2-sin x))^(3/2)  
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Basic functions

1. What is the range of k(x)=-|x| Asking for the range is another way of asking for valid Y values (or in this case k(x) values). So the range is (-infinity, 0]. 2.Is y=2 a constant function? Yes. Constant functions have no variation in the output for any input. 3. x=-3 a function? No it is not. In order to be a function, there are several properties which must be satisfied, one of which is that for any input, there must be a single output. Often we simplify this by asking "does the function pass a vertical line test?" So if you are to draw a vertical line, would it only cross the function in one place for all values of X? For X=-3, it is a vertical line, so it fails this property of functions because there are an infinite number of solutions (Y values) for this single X value (-3). 4. What is the minimum value of y=x^2-4 The minimum value of parabola's will occur either at the vertex (where its slope is 0) or at its domain boundaries. There are no stated domain restrictions for this function and the slope is 0 at X=0. The first derivative is y'=2x and y'=0 at x=0. The second derivative is y''=2, indicating it is concave for all values X. So the minimum is x=0 and y=-4. For a function with a higher power, there may be more than one minimum (each called local minimum) and any place the second derivative is concave up, both to the left and right segments where the first derivative is 0 will be one of these local minimums. You will take your inputs for each relative minimum and any domain limits (for functions which do not have all real numbers for their domain) and compare the Y values of the original function to determine the absolute minimum for that function.
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second derivate of sqrt(x²+1)

Given sqrt(x²+1)   First derivative of the equation, d/dx= x/√(x^2+1)   Again derivative of the equation d^2/dx^2=1/(x^2+1)^(3/2) Get Free Calculus Help  
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help me!! :(

Let's look at each answer in turn. A. It can't be the answer. The reason is that the constant amount increases the wage year on year, so the wage is getting bigger but the increment is becoming a smaller fraction, or percentage, of the wage as time goes on. B. This can't be the answer even though the increment itself is increasing. The percentage increase isn't in step with this. 0.50/13.50=3.7% approx. The wage is $14/hr. Next year the percentage is 0.75/14=5.4%, so the percent increase is not constant. C. This is the correct answer because the wage rises by $2, which is 2/20=10% increase. Next year the increase percent is 2.20/22=10%. That's a constant percent increase. D. This can't be correct if C is correct. Let's see what we've got. 0.75/15=5%; 1.75/17=10.3%. Not a constant increase.
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Second Derivative

Given f"(x) = 12x^3+12x^2 Differentiate with respect to x 3*12x^2+2*12x 36x^2+24x Differentiate with respect to x 36*2x+24 72x+24 Get detailed explanation on Second Derivative
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limit trigonometry

(tanx-x)/(x-sinx)=(sinx-xcosx)/(xcosx-sinxcosx). As x approaches 0 the expression approaches 2. Why? To find out, let's use some calculus. Let's suppose that sinx=a[0]+a[1]x+a[2]x^2+a[3]x^3+...+a[n]x^n, a polynomial series in x, with real coefficients a[0], etc. We are going to use a value for x<1 so the series would converge. When x=0, sinx=0 so a[0]=0. The derivative of sinx is cosx, so cosx=a[1]+2a[2]x+3a[3]x^2+...+na[n]x^(n-1). When x=0 cosx=1, so a[1]=1 and sinx=x+a[2]x^2+... The second derivative of sinx is -sinx=2a[2]+6a[3]x+... When x=0, sinx=0, so 2a[2]=0 and a[2]=0 and sinx becomes x+a[3]x^3. Moving on to the next derivative which is -cosx we have 6a[3]+... But -cosx when x=0 is -1, so a[3]=-1/6, making sinx=x-x^3/6+... We could keep on going through more derivatives, but it's not necessary because we can establish right now what x-sinx tends to as x approaches zero; x-sinx=x^3/6. Now, let tanx=b[0]+b[1]x+b[2]x^2+... just like we did with sinx. We can see that b[0]=0 because tanx=0 when x=0. The first derivative of tanx=sec^2x=b[1]+2b[2]x+3b[3]x^2+... secx is the reciprocal of cosx which is 1 when x=0, so b[1], like a[1], is 1 and tanx=x+b[2]x^2. The derivative of sec^2x is the derivative of cos^-2x=-2cos-3x*(-sinx)=2tanxsec^2x=0, when x=0, making b[2]=0. We need to go to the next derivative to find b[3]. But 2tanxsec^2x=2tanx(1+tan^2x)=2tanx+2tan^3x, which differentiates to 2sec^2x+6tan^2xsec^2x, and this derivative is 2 when x=0. Therefore 2=6b[3] and b[3]=1/3 and tanx-x=x^3/3. So the limit as x approaches zero of (tanx-x)/(x-sinx)=(x^3/3)/(x^3/6)=2.  
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