Review your knowledge of the Quotient rule for derivatives, and use it to solve problems.
Log in J Kuijpers 7 years agoPosted 7 years ago. Direct link to J Kuijpers's post “In my studybook from coll...” In my studybook from college I've found this problem: • (17 votes) Ian Pulizzotto 7 years agoPosted 7 years ago. Direct link to Ian Pulizzotto's post “There's even a quicker wa...” There's even a quicker way! We can first simplify the expression inside the logarithm, by factoring out e^x in the numerator: (61 votes) Scott Thomas Carter 6 years agoPosted 6 years ago. Direct link to Scott Thomas Carter's post “What would be the reason ...” What would be the reason that one couldn't simply take a quotient (for example (x^2 - 3)/x^4 and simply make it the product (x^2-3)*x^-4 and then apply the product rule? Of course, they have different answers because the product rule does not factor in the square of the denominator, but what rule would be broken by doing so? • (2 votes) Ian Pulizzotto 6 years agoPosted 6 years ago. Direct link to Ian Pulizzotto's post “Yes, you can express (x^2...” Yes, you can express (x^2 - 3)/x^4 as the product (x^2 - 3) * x^-4 and use the product rule to take the derivative. No rule is broken here. Your answer might not appear the same as if you used the quotient rule to differentiate (x^2 - 3)/x^4, but it should end up mathematically equivalent. If not, then you've made an algebraic or differentiation error somewhere. In fact, the product and chain rules can be used to derive the quotient rule: d/dx of [f(x)/g(x)] = d/dx of {f(x)*[g(x)]^(-1)} Have a blessed, wonderful day! George Emmanuel 5 years agoPosted 5 years ago. Direct link to George Emmanuel's post “should it not be written ...” should it not be written like this [d/dx(g(x)) .f(x)- • (0 votes) Alex 5 years agoPosted 5 years ago. Direct link to Alex's post “Your textbook is incorrec...” Your textbook is incorrect; from all other sources, it is indeed f'g - g'f. Try deriving it yourself using the product rule and chain rule. (11 votes) Daleks Incorporated a year agoPosted a year ago. Direct link to Daleks Incorporated's post “For the AP test, do you n...” For the AP test, do you need to fully simplify your answers like you require in the review questions listed here? • (4 votes) Angelo Luidens 7 years agoPosted 7 years ago. Direct link to Angelo Luidens's post “In the first Checking You...” In the first Checking Your Understanding above, isn't the d/dx of e^x = xe^(x-1) ? Else what is the rule applied to e^x? • (2 votes) Howard Bradley 7 years agoPosted 7 years ago. Direct link to Howard Bradley's post “The power rule only appli...” The power rule only applies when the exponent is a constant. For example x³ or x ⁻². There's quite a long section on the derivatives of exponential functions, here's the video on eˣ (4 votes) mwood21 3 years agoPosted 3 years ago. Direct link to mwood21's post “For question 1 I used x^2...” For question 1 I used x^2 as my f(x) instead of e^x and ended up getting [e^x(2-x)]/(x^3). Would this be incorrect? • (1 vote) Hecretary Bird 3 years agoPosted 3 years ago. Direct link to Hecretary Bird's post “Division isn't commutativ...” Division isn't commutative like multiplication, so if you switch the positions of the numbers you're dividing, you'll get a different answer. From this, it follows that the derivative of one function divided by a second one would be different than the derivative of the second divided by the first. You don't have to be careful about this when doing the product rule, but when doing the quotient rule, remember that you subtract term with the derivative of the bottom function, and divide by the bottom function squared. Anything else shouldn't give you the right answer, and (e^x (2-x)) / x^3 would be incorrect. (3 votes) kris kirk 5 years agoPosted 5 years ago. Direct link to kris kirk's post “so if i had for example '...” so if i had for example '''y=200x/3+5x^2''' and had to differentiate with respect to y, how would i go about this? ( im pretty certain it's quotient rule) i THINK i have it but i also think i've done it with respect to x though... • (1 vote) ggadget6 5 years agoPosted 5 years ago. Direct link to ggadget6's post “No quotient rule required...” No quotient rule required :). You just need the normal derivative rules. Since there are no x's in the denominator, only constants, you can treat 200/3 as a constant, and just use the normal power rule. In this case, your answer would be dy/dx = 200/3 + 10x. (2 votes) kevinclauson99 8 years agoPosted 8 years ago. Direct link to kevinclauson99's post “In problem 2 of example 1...” In problem 2 of example 1 you state that the answer is (−xsin(x)+3cos(x))/x^4 however the actual answer is (which the site recognizes) is flip out that middle "+" for a "-" • (1 vote) Jesse Cooper 8 years agoPosted 8 years ago. Direct link to Jesse Cooper's post “The first minus sign in t...” The first minus sign in the solution has been brought out the front of the fraction, so they are stating the answer is −(xsin(x)+3cos(x))/x^4, which is equal to (−xsin(x)-3cos(x))/x^4, rather than (−xsin(x)+3cos(x))/x^4 (2 votes) muhammad muttayab 2 years agoPosted 2 years ago. Direct link to muhammad muttayab's post “can rational function be ...” can rational function be differentiated by doing partial function ? if yes,then which one is more preferable partial fraction or quotient rule? • (1 vote) Malone Jacoway 2 years agoPosted 2 years ago. Direct link to Malone Jacoway's post “A rational function can b...” A rational function can be split into partial fractions before taking the derivative, but this is often a more lengthy process than just doing the quotient rule. (1 vote) kaitlynmarushia95 5 years agoPosted 5 years ago. Direct link to kaitlynmarushia95's post “5x^2/x^2 -9Why is the 5...” 5x^2/x^2 -9 Why is the 5 put to the side when getting f'(x) But the correct answer is -90x/(x^2 -9)^2 I hope someone can help explain this to me. • (1 vote) Timur Kozhamuratov 5 years agoPosted 5 years ago. Direct link to Timur Kozhamuratov's post “If we set 5x^2 = g(x) (fo...” If we set 5x^2 = g(x) (for example) (1 vote)Want to join the conversation?
differentiate ""ln((1+e^x)/(1+e^(-x)))""
and I don't seem to be able to crack it, how kan I solve this one?
(1+e^x)/(1+e^(-x)) = (e^x)(e^(-x)+1)/(1+e^(-x)) = (e^x)(1+e^(-x))/(1+e^(-x)) = e^x.
Therefore, we have
d/dx of (ln((1+e^x)/(1+e^(-x)))) = d/dx of (ln(e^x)) = d/dx of (x) = 1. Amazing!
= f(x)*(-1)[g(x)]^(-2)*g'(x) + f'(x)*[g(x)]^(-1)
= [g(x)]^(-2) * [-f(x)*g'(x) + f'(x)*g(x)]
= [g(x)*f'(x) - f(x)*g'(x)]/[g(x)]^2.
d/d(x)(f(x)).g(x)]/[g(x)]^2, i found this lesson extremely baffling as this opposes what i learned from my textbook, which formula is the right one? any suggestion to improve my understanding,please?
It does not apply when the exponent is a variable. For example 2ˣ or, as in this case eˣ.
https://www.khanacademy.org/math/ap-calculus-ab/ab-derivatives-advanced/ab-diff-exp/v/derivative-of-ex
the answer I continue to get is 10x(10x^2 +10x-90)/ (x^2 -9)^2
and x^2-9 = h(x)(for example)
g'(x) = 5 (2x^2-1)= 10x
h'(x) = 2x (because h'(x)= 2x^2-1 and 9 =0 according to "power rule",
which you can check out on Khan academy).
then f'(x)=g'(x)/h'(x)
= (g'(x)*h(x)-g(x)*h'(x))/g(x)^2
substitute (10x)(x^2-9)-(5x^2)(2x)/(x^2-9)^2
= 10x^3-90x-10x^3/(x^2-9)^2
= -90x/(x^2-9)^2
Hope that helps...;)
