How do I find the derivative of y=1/(ax-b)?
You apply the ruleso you will haveVerification
What is the derivative of ln(x^2+y^2)?
You should use substitution in this case. If a problem seems difficult, try to simplify it for yourself. Well we know from the basic derivation of ln u is: (ln u)' = u' * (1/u) Ex: (ln 2x)' = 2 * (1/2x) = 1/x Using that as your basis, you should be able to solve the problem.
Find the derivative y=e^5x?
The derivative of e^x is just e^x dx, the same thing. but when you have something else in the exponent you have to apply the chain rule. So it's the derivative of e^(5x) times the derivative of 5x. This gives you y' = 5e^(5x). I'm assuming here that you meant e^(5x) and not (e^5)x. If it's the latter, then yes, the derivative would just be e^5 because this is just a constant times x.
Find the derivative of y=x^cos(x)?
I'm having a bit of trouble with this one. Here is what I'm doing: 1) Write everything in terms of ln ln(y)=cos(x)*ln(x) 2) Differentiate (1/y)=cos(x)*(1/x) Since cos(x) is a constant, shouldn't it just come along without being differentiated? 3) Distribute the y y`=(x^cos(x))*(cos(x)*(1/x)) Shouldnt that be it? What am I doing wrong?
Find the derivative of y=tan^-1 √3x?
For this problem, lets use the chain rule: "take the derivative of the outside with the inside intact, times the derivative of the inside" Remember that the derivative of tangent inverse is: 1/(1+x^2) so: dy/dx = 1 / (1 + 3x) * d/dx (3x)^1/2 = 1/(1 + 3x) * .5(3x)^(-1/2) * d/dx(3x) = 1/(1 + 3x) * .5(3x)^(-1/2) * 3 Now simplify = 3 / [ (1 + 3x) 2root(3x) ]
How do I find the derivative of x^sin x?
Because both x & sin(x) are functions of x, we cannot really use the derivative rule for powers, because that only works when the exponent is a constant (or independent of x).We have to get a little creative about this:Let's say: [math]y = x^{sin(x)}[/math]This would be much simpler if x & sin(x) were separated and not connected exponentially. What makes exponents "simpler"?Taking the natural logarithm of both sides:[math]ln(y) = sin(x)ln(x)[/math]Now we finally have something we can work with. The left side is a not-so-bad function of y, and the right hand side is the product of two functions of x whose derivatives are known (in case you don't know, please go back to calculus class :) ). Taking the derivative of both sides: [math]\frac{d ln(y)}{dx} = sin(x) * \frac{1}{x} + cos(x)*ln(x)[/math]By the chain rule, we can differentiate ln(x):[math]\frac{1}{y} * \frac{dy}{dx} = sin(x) * \frac{1}{x} + cos(x)*ln(x)[/math][math] \frac{dy}{dx} = y * (sin(x) * \frac{1}{x} + cos(x)*ln(x))[/math]Substituting y back into the equation:[math] \frac{dy}{dx} = x^{sin(x)} * ( \frac{sin(x)}{x} + cos(x)*ln(x))[/math]This method of differentiation is similar to Implicit differentation .
What is the derivative of y=cos(x+y)?
We have [math]y=[/math][math]cos(x+y)[/math]Differentiating,[math] \dfrac{dy}{dx} =−sin(x+y)(x+y)′[/math] [Chain rule][math]⇒\dfrac{dy}{dx}=−sin(x+y)(1+\dfrac{dy}{dx})[/math]Rewriting,[math]\dfrac{dy}{dx}=−sin(x+y)−sin(x+y)⋅\dfrac{dy}{dx}[/math][math]⇒\dfrac{dy}{dx}+sin(x+y)\dfrac{dy}{dx}=−sin(x+y)[/math]Taking [math]\dfrac{dy}{dx}[/math] common,[math]\dfrac{dy}{dx}(1+sin(x+y))=−sin(x+y)[/math]Bringing [math](1+sin(x+y))[/math] to other side,[math]\dfrac{dy}{dx}=−\dfrac{sin(x+y)}{1+sin(x+y)}[/math] will be the final answer.
What is the derivative of y=tan(x+y)?
[math]y = tan(x + y)[/math]Extracting inverse on both sides,[math]tan^{-1}y = x + y[/math]Differentiating both sides,[math]\frac{1}{1+y^2}\cdot\frac{dy}{dx} = 1 + \frac{dy}{dx}[/math][math]\big(\frac{1}{1+y^2} - 1\big)\cdot\frac{dy}{dx} = 1[/math][math]\frac{1-(1+y^2)}{1+y^2}\cdot\frac{dy}{dx} = 1[/math][math]\big(\frac{-y^2}{1+y^2}\big)\cdot\frac{dy}{dx} = 1[/math]Making [math]\frac{dy}{dx}[/math] as the subject of the formula,[math]\frac{dy}{dx} = -\big(\frac{1+y^2}{y^2}\big)[/math]Replacing [math]y = tan(x+y),[/math][math]\frac{dy}{dx} = -\big(\frac{1+tan^2({x+y})}{tan^2({x+y})}\big)[/math][math]\frac{dy}{dx} = -\csc^2({x+y})[/math]
Finding the derivative of Y?
Since we are taking the derivative of y IN TERMS OF X, we have to use chain rule for the y^2. That y^2 has a whole bunch of x's in that we do not know of. For example, let's say (randomly) that y turned out to be (2-x). Then, to take this derivative of y^2, we would have to use power rule then take the derivative of the inner function (2-x). For the 5xy part, use the product rule. The product rule is taking the first term (x) and multiplying it by the derivative of the second term (y), plus the second term (y) and multiplying it by the derivative of the first term (x). 5 is a constant, so we don't have to worry about that. So, we get 5 (x dy/dx + y dx/dx). Which equals 5 (x dy/dx + y). Once you get this, move all the terms with dy/dx to one side of the equation and move all the terms without dy/dx to the other side. Then, on the side with the terms with dy/dx, factor out a dy/dx. Then, divide both sides by the factor. See below: 2x + 2y (dy/dx) = 5x dy/dx + 5y 2x - 5y = 5x dy/dx - 2y (dy/dx) 2x - 5y = dy/dx (5x - 2y) Divide both sides by 5x - 2y (2x - 5y) / (5x - 2y) = dy/dx Factoring out a -1 in both the numerator and denominator will give you that book answer. The -1 factored out in both places will cancel out.
How can I find the 50th derivative of y=cos2x?
Trig function derivatives are cyclical and repeat every four derivatives (so sin''''(x) = sin(x), etc). You can divide 50 by four and take the remainder, since every four derivatives will simply cancel out. The remainder in this case is 2. The second derivative of cos(x) is -cos(x).Now for the coefficient. Every time you take a derivative, the coefficient on the term inside the trig function (2 in this case) comes out and multiplies the whole function. So the coefficient of cos'(2x) would be 2, the coefficient of cos''(2x) is 4, etc. You are doing this 50 time so the coefficient would be 2^50.Put this together and you get 2^50(-cos(2x))