How to differentiate y = (x^2 - 2x)/sqrt(x)?
OMG, i actually managed to work it out, never thought i could anyway, here it goes. u turn what u have into this: (x^2 - 2x) / x^1/2 = (x^2 - 2x) * x^-1/2 = x^3/2 - 2x^1/2 Now, u have that, adn so now u can differentiate it because its a nice simple expression (no fractiosn and all that. u do the dy/dx thing and u get this. 3/2*x^1/2 - 1/2*2x^-1/2 = (3x^1/2)/2 - x*-1/2 = (3x^1/2)/2 - 1/(x^1/2) = (3x^1/2)/2 - 1/sqrt(x) Now u use common denominators = [sqrt(x)*(3x^1/2) - 2] / 2sqrt(x) So now u basically just multiply sqrt(x) by 3 times sqrt(x), which i wrote as 3x^1/2, ad u get. = 3x - 2 / 2sqrt(x) tadaaa, there u go, im rlly new to calculus, so this is pretty big for me :P
Can you differentiate (x^2+4x+3)/(sqrt(x))?
differentiate making use of product rule etc. dy/dx(x^2 + 4) + y(2x) = 4 + (a million/y)(dy/dx) assemble words with dy/dx onto one side and factor dy/dx(x^2 + 4 - a million/y) = -2xy + 4 dy/dx = -2(xy - 2) / (x^2 + 4 - a million/y) Now you ought to bypass away it like this, or simplify added via turning the denominator into one term. even however you ought to desire in terms of merely x, yet its puzzling simply by lny term,.. i don't understand if which would be obtainable or not.
How do you differentiate ((x^2 + 4x +3)/(sqrt x))?
Use the quotient rule. f(x) = x^2 + 4x + 3 g(x) = x^0.5 f'(x) = 2x+4 g'(x) = 0.5*x^-0.5 Quotient rule: [f(x)/g(x)]' = (f'g - g'f)/g^2 So, this becomes: [(2x+4)*x^0.5 - 0.5*(x^-0.5)*(x^2+4x+3)]/(x^0.5)^2 Let's see if we can simplify: Denominator is simply x. that's easy. --------------------------------- Numerator is: [(2x+4)*x - 0.5*(x^2+4x+3)]/x^0.5 (Notice that I multiplied the first term by (x^0.5)/(x^0.5) (2x^2+4x - 0.5x^2 - 2x -1.5)/x^0.5 (1.5x^2 - 2x - 1.5)/x^0.5 ------------ Remember that this is just the numerator. The denominator is x so the derivative of f/g is: (1.5x^2 - 2x - 1.5)/x^1.5 Edit: Please check my arithmetic to be sure I don't have a typo.
How do you differentiate [math](\sqrt{x})^{\sqrt{x}}[/math]?
How do you differentiate [math](\sqrt x)^{\sqrt x}[/math]?Everyone who has studied calculus knows these two rules (they are basic rules, combined with an application of the chain rule):[math]\frac{d}{dx}\left(f(x)^b\right)=\color{red}{bf(x)^{b-1}\cdot f'(x)}[/math] (a function raised to a constant power)[math]\frac{d}{dx}\left(a^{g(x)}\right)=\color{blue}{a^{g(x)}\ln a\cdot g'(x)}[/math] (a constant raised to the power of a function)There’s a lesser known rule that applies when both the base and the exponent are functions (nonconstant); it is basically just the sum of the two rules above:[math]\frac{d}{dx}\left(f(x)^{g(x)}\right)=\color{red}{g(x)f(x)^{g(x)-1}\cdot f'(x)} + \color{blue}{f(x)^{g(x)}\ln f(x)\cdot g'(x)}[/math]We can take out the common factor of [math]f(x)^{g(x)-1}[/math], leaving[math]\boxed{\frac{d}{dx}\left(f(x)^{g(x)}\right)=f(x)^{g(x)-1}\left(g(x)f'(x) + f(x)g'(x)\ln f(x)\right)}[/math]In this particular case, with [math]f(x)=g(x)=\sqrt x[/math], we have [math]g(x)f'(x)=f(x)g'(x)=\frac{1}{2}[/math], so this simplifies to[math]\frac{d}{dx}\left({\sqrt x}^{\sqrt x}\right)=\frac{1}{2}\sqrt x^{\sqrt x-1}\left(1 + \frac{1}{2}\ln x\right)[/math]The general rule (which can be proven using the logarithmic differentiation methods described in many of the other answers) can be used to make short work of problems like:[math]\frac{d}{dx}\left((\cos x)^{\sin x}\right)=(\cos x)^{\sin x-1}\left(-\sin^2x + \cos^2x\cdot\ln\sin x\right)[/math][math]\frac{d}{dx}\left((x^2+1)^{(e^x)}\right)=(x^2+1)^{(e^x-1)}\left(2xe^x + x^2e^x\ln(x^2+1)\right)[/math]
Differentiate x^3 -4x+6?
all you have to do is make the equation one degree less x^3 = x^2, x^4 = x^3 and so on Therefore everything else is one degree smaller -4x becomes -4, +6 becomes 0 If there is an exponent, put it in front of x or multiply by what is in front of x. I am dead serious watch this video it will explain so much. It did for me. http://www.thinkwell.com/a/calculusin20minutes?utm_source=youtube&utm_medium=info&utm_campaign=calcin20min
How do I differentiate X2 + 3?
f(x) = x^2 + 3Differentiatingd/dx(x^2) + 0Because differentiation of a constant is 0.2x^(2–1)Because differentiation of x^n is nx^n-1So the answer is 2x
How do I differentiate √1-x^2?
[math]\dfrac{\mathrm d}{\mathrm dx} \sqrt{1-x^2}[/math][math]= \displaystyle \lim_{h\to0} \dfrac {\sqrt{1-(x+h)^2} - \sqrt{1-x^2}} {h}[/math][math]= \displaystyle \lim_{h\to0} \dfrac {(\sqrt{1-(x+h)^2} - \sqrt{1-x^2})(\sqrt{1-(x+h)^2} + \sqrt{1-x^2})} {h(\sqrt{1-(x+h)^2} + \sqrt{1-x^2})}[/math][math]= \displaystyle \lim_{h\to0} \dfrac {\sqrt{1-(x+h)^2}^2 - \sqrt{1-x^2}^2} {h(\sqrt{1-(x+h)^2} + \sqrt{1-x^2})}[/math][math]= \displaystyle \lim_{h\to0} \dfrac {(1-(x+h)^2) - (1-x^2)} {h(\sqrt{1-(x+h)^2} + \sqrt{1-x^2})}[/math][math]= \displaystyle \lim_{h\to0} \dfrac {x^2-(x+h)^2} {h(\sqrt{1-(x+h)^2} + \sqrt{1-x^2})}[/math][math]= \displaystyle \lim_{h\to0} \dfrac {-2xh-h^2} {h(\sqrt{1-(x+h)^2} + \sqrt{1-x^2})}[/math][math]= \displaystyle \lim_{h\to0} \dfrac {-2x-h} {(\sqrt{1-(x+h)^2} + \sqrt{1-x^2})}[/math][math]= \displaystyle \dfrac {-2x} {\sqrt{1-x^2} + \sqrt{1-x^2}}[/math][math]= \displaystyle \dfrac {-x} {\sqrt{1-x^2}}[/math]
how do I differentiate. Y=(4x^4) - (6x^3) + (2x^2)- 1?
Define the function you've given as [math]f(x)=4x^4-6x^3+2x^2-1[/math]. You can determine the derivative of [math]f(x),[/math] which we denote by [math]f'(x)[/math] by using the limit definition of the derivative or the conventions mentioned by Terry Moore. I'll do this both ways. (1) Limit Definition of the Derivative The limit definition of the derivative is:[math]f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}[/math] Applying this to our function [math]f(x)[/math] we obtain the following:[math]\begin{align*}f'(x)&=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\\&=\lim_{h\to 0}\frac{4(x+h)^4-6(x+h)^3+2(x+h)^2-1-(4x^4-6x^3+2x^2-1)}{h}\\&=\lim_{h\to 0}\frac{4h^4+16h^3x-6h^3+24h^2x^2-18h^2x+2h^2+16hx^3-18hx^2+4hx}{h}\\&=\lim_{h\to 0}\frac{h(4h^3+16h^2x-6h^2+24hx^2-18hx+2h+16x^3-18x^2+4x)}{h}\\&=\lim_{h\to 0}(4h^3+16h^2x-6h^2+24hx^2-18hx+2h+16x^3-18x^2+4x)\\&=16x^3-18x^2+4x\end{align*}[/math]As you can see, using the limit definition to compute derivatives is timely and not very efficient. (2) Using ConventionsThe faster way that every calculus student uses to compute derivatives is to memorize a set of derivative rules (again, those mentioned by Terry Moore) to speed up computations. The rule for powers of [math]x[/math] is to simply carry down the exponent, multiply it by the term, and decrease the power of [math]x[/math] by one. For example, if [math]g(x)=x^3[/math] the derivative is [math]g'(x)=3x^{3-1}=3x^2.[/math] Also, note that the derivative of any constant is just 0. That is, if [math]h(x)=4[/math] then [math]h'(x)=0[/math]. We can easily check that the derivative of [math]f(x)[/math] we compute using these rules is the same as that given by the limit definition:[math]\begin{align*}f'(x)&=4(4)x^{4-1}-6(3)x^{3-1}+2(2)x^{2-1}-0\\\implies f'(x)&=16x^3-18x^2+4x\end{align*}[/math]which is exactly what we obtained above.
Differentiate the following function. y= (7x^2 + 3 sqrt x)/ 2x?
The first thing to do is to change sqrt(x) to (x^(1/2)) Then you apply the quotient law of differentiation: d(u/v) dx = [(v * du/dx) - (u * dv/dx)] / v^2 ------------------ y = [(7*(x^2)) + (3*(x^(1/2)))] / 2x u = {7*(x^2) + [(3 * (x^(1/2)))]} du/dx = 14x + [(3) * (1/2) * (x^(-1/2))] = [14x + (3 / (2 * sqrt(x)))] v = 2x dv/dx = 2 v * du/dx = 2x * [14x + (3 / (2 * sqrt(x)))] = [28*x^2 + [3 * sqrt(x)) / 2] u*dv/dx = 2 * {7*(x^2) + [(3 * (sqrt(x)))]} = {14*(x^2) + [(3 * (sqrt(x))]} v^2 = 4x^2 Now that the basic calculations are done, finishing the problem is just a matter of combining the three expressions (v*du/dx, u*dv/dx, and v^2) and simplifying. However it looks like the algebraic manipulations could be messy. ----------------- So the first step is just the combining part (I'm copying these through the clipboard so that I won't make a mistake copying by typing.) [28*x^2 + [3 * sqrt(x)) / 2] - {14*(x^2) + [(3 * (sqrt(x))))]} / [(4x^2)].....<<<...Answer Since (sqrt(x) / (x^2)) = x^(-3/2) and x^2/ x^2 = 1, that can be simplified. Trying to divide by 4 would just make things worse. So this is equal to: {[28 + ((3/2) * (x^(-3/2)))] - [(14 + (3 * (x^(-3/2))))]} / 4 = {[(14 - (3/2) * (x^(-3/2)))] / 4}....<<<...Answer in simplified form .
How would you differentiate the equation [math]x^2 + 4xy + y^2=20[/math]?
The given equation is x² + 4xy + y² = 20On differentiating w. r. t. xwe get, d/dx( x² + 4xy + y² ) = d/dx(20) 2x + (4x)(dy/dx) + 4y + (2y)(dy/dx) = 0[ On apply product rule ( Ist function.derivatives of IInd function + IInd function.derivative of Ist function ) on the middle term (4xy) & differentiation of constant (20) is zero]or, 2x + 4y + ( 4x + 2y )(dy/dx) = 0 (4x +2y)(dy/dx) = -(2x + 4y) dy/dx = - ( 2x + 4y )/( 4x + 2y)or, dy/dx = - ( x + 2y )/(2x + y). Answer.