How do i check to see if a derivative I found is correct?

Question

In calculus, is there a way to see if the derivative and the second derivative are correct either by graphing or using the calculator? i can find the derivative rather easily but is there a way to make sure i did it correctly?please help

Answer 1

If you take the integral, it should produce the original equation with a constant added.

Answer 2

Usually if you have a graphing calculator, you can check the derivative of a function at a certain point.

Say you have x^3 + 5 = f(x)

Plug that into your calculator, and you know the derivative of this function is 3x^2, but if you plug in 2, you'll get 12. Now on you calculator, just use the derivative tool on the graph and it should say 12. This is found on all of the TI graphing calcs, and TI-89's actually find the derivative of a function you input!

Answer 3

you're headed contained in the wonderful suited path. you ought to use ln to locate dy/dx. So... ln(y) = xln((2^xtanx)/(e^x)) From the way which you typed the equation, i'm assuming that the "tanx" area of the equation isn't in the two^x exponent. Now we take the spinoff. y'/y = ln((2^xtanx)/(e^x)) + x(e^x/(2^xtanx))(((2^xln(2)tanx + 2^xsecxtanx)(e^x)-((2^xtanx)/e^x)/((e^x)... i think of it rather is right. might have miscounted some brackets. Anways, first, use the product rule on the "x" and "ln" areas. Then as you're taking the spinoff of the "ln" area, first you're taking the ln spinoff of what's contained in the brackets (that's one million/regardless of), then the spinoff of what's contained in the ln brackets (quotient rule). then you definately multiply the two facet by means of "y," that's the unique equation. Your answer, the spinoff, is the long equation above expanded by means of y. y' = ((2^x tanx)/e^x)^ x)((ln((2^xtanx)/(e^x)) + x(e^x/(2^xtanx))(((2^xln(2)tanx + 2^xsecxtanx)(e^x)-((2^xtanx)/e^x)/((e^x)... each and every so often you're caught with very long spinoff equations in calculus and there is not any longer something which you're able to do approximately it, which sucks. I completely comprehend your frustrations. desire this facilitates :) P.S. as quickly as I submitted this, i observed that the spinoff equation ended with e..., which will desire to be ((e^x)^2) (that's the denominator squared).

Answer 4

Take the integral of it, and you should get function you started with.