anonymous
  • anonymous
What is the derivative: y=ln(x^5sinxcosx)
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
do you mean \[\LARGE y=\ln (x^{5\cdot \sin x \cdot \cos x}) \] OR \[\LARGE y=\ln(x^5 \cdot \sin x \cdot \cos x)\] ??
anonymous
  • anonymous
Well, you pretty much should have all the derivatives down pat by now. The only thing I haven’t gone into is trigonometric functions. I figured I’d make everything complicated after I’m done with the formulas. Incidentally, if you don't feel comfortable with the material yet, I have problem sets linked at the end of every section. I don't know if I've made that clear. The derivates of all trig functions are in the reference table. However, I am just going to work with 3 basic trig derivatives here. y = sin x y’ = cos x y = cos x y’ = -sin x y = tan x y’ = sec2 x These are rules that you will have to follow. They are also very basic, and can be complexified in interesting ways. (Did I just create a word?) In the spirit of having you actually remember some of the above, I will allow you to see how the derivative of sin x is cos x. Carefully draw both a sin and cos curve on one graph, from 0 to 2∏. (That symbol is pi) You will notice at x = 0 that the angle that the sin curve goes through the origin is a 45 degree slop upwards, which means that it goes up 1 y for every 1 x. The slope is then 1, and the derivative at that point should be 1. Cos of 0 is 1. At x = ∏/2, sin x reaches a maximum, at which the slope is 0. Cos ∏/2 is 0. At ∏, the sin curve has a -1 slope, and cos is -1. This pattern is correct not only at these points but at every point. A similar analysis will show that the derivative of cos x is -sin x. With complex rules y = 5sin x y’ = 5cos x This is because the 5 is a coefficient, and any time you are taking a derivative of a function with a number in front, you can take that number away for the purposes of taking the derivative, and put it back in after solving. y = sin 5x y’ = 5cos 5x This is simply the chain rule, used in the last topic. The 5x is u and the derivative of sin u is cos u. The derivative of 5x is 5. Multiply the derivative of outside by derivative of inside, and you have the derivative of equation. Then you replace the u. y = 3sin 5x y’ = 15cos 5x y = 3cos 5x y’ = -15sin 5x y = sin2 x = (sin x)2 = u2 (and u = sin x) y’ = 2u(cos x) = 2sin x cos x This one I solved by a chain rule similar to what you saw in the chain rule section. Notice that I put the square around the entire sin x, this is explained in the Intro. Now I made sin x the inner function, and u2 the outer function. The rest is simple. Take the derivatives of both and multiply. Then replace the u to its original status (sin x). y = 3sin2 (5x + 3) This one is not as tough as it looks if you understood everything before. On the other hand, if you understood everything before it probably doesn’t look too tough. This requires a chain rule inside a chain rule. The equation rearranges as 3[sin (5x + 3)]2 The outside equation to solve is u2 and u = sin (5x +3). So If I take the derivative of the outside, getting 2u, and multiply it by the derivative of the inside function, then I have my answer. But what is the derivative of the inside function? The derivative of sin (5x +3) requires a chain rule. Once you figure out the derivative, take it back and multiply it by the other one. The derivative of this is 5 * cos (5x + 3). Multiply that by 2u, and multiply by the 3 that we've been ignoring, and you have 30u cos (5x + 3) which is 30[sin (5x + 3)]cos (5x + 3). I hope I didn’t screw that up. No, I think it’s right. y = 2tan4 x y’ = 8tan3 x sec2 x This is nothing you can’t handle. The u turned back into a tan x at the end, and 8u3sec2x became 8tan3 x sec2 x. y = sin (ln x) y’ = cos (ln x) * 1/x y = ln (sin x) y' = (1/sin x) * (cos x) = cos x/sin x = cot x y = ln (cos x) y' = (1/cos x) * (- sin x) = - sinx/cos x = - tan x y = cos (ln 2x) y’ = -sin (ln 2x) * 1/x I used two chains in that one. The 2x must become a u in order to get the derivative of a ln 2x. The following example uses a derivative that you are required to know, and is shown along with the rest in my reference tables. y = tan x/sin x = sec x y' = sec x tan x y = tan (exx2) y’ = sec2 (exx2) * (exx2 + ex2x) That one has a product rule inside the chain rule. Make sure your sec2 is left with a (exx2) inside, because during the solving process it turns into a u. OK, I think I’ve done enough trigonometries for one day. I’ll move on to implicit differentiation.
anonymous
  • anonymous
First Derivative Formulas y is a function y = y(x) C = constant, the derivative(y') of a constant is 0 y = C => y' = 0 example: y = 5, y' = 0 If y is a function of a type y = xn the formula for the derivative is: y = xn => y' = nxn-1 example: y = x3 y' = 3x3-1 = 3x2 y = x-3 y' = -3x-4 From the upper formula we can say for derivative y' of a function y = x = x1 that: if y = x then y'=1 y = f1(x) + f2(x) + f3(x) ...=> y' = f'1(x) + f'2(x) + f'3(x) ... This formula represents the derivative of a function that is sum of functions. example: If we have two functions f(x) = x2 + x + 1 and g(x) = x5 + 7 and y = f(x) + g(x) then y' = f'(x) + g'(x) => y' = (x2 + x + 1)' + (x5 + 7)' = 2x1 + 1 + 0 + 5x4 + 0 = 5x4 + 2x + 1 If a function is multiplication of two functions the derivate formula is: y = f(x).g(x) => y' = f'(x)g(x) + f(x)g'(x) If f(x) = C(C is a contstant), and y = f(x)g(x) y = Cg(x) y'=C'.g(x) + C.g'(x) = 0 + C.g'(x) = C.g'(x) y = Cf(x) => y' = C.f'(x) There are examples of the following formulas in the task section. y = f(x) g(x) y' = f'(x)g(x) - f(x)g'(x) g2(x) y = ln x => y' = 1/x y = ex => y' = ex y = sin x => y' = cos x y = cos x => y' = -sin x y = tan x => y' = 1/cos2x y = cot x => y' = -1/sin2x y = arcsin x => y' = 1 √1 - x.x y = arccos x => y' = -1 √1 - x.x y = arctan x => y' = 1 1 + x2 y = arccot x => y' = -1 1 + x2 if a function is a function of function: u = u(x) y = f(u) => y' = f'(u).u'

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
\[y=\ln(x^5sinxcosx)\]
anonymous
  • anonymous
|dw:1333105215516:dw||dw:1333105357359:dw||dw:1333105592937:dw| hmmm. idk...

Looking for something else?

Not the answer you are looking for? Search for more explanations.