I'm very bad at these graph sketching questions. Any suggestions for where I can practice such questions from? :P
I'm not that good at those either to be honest. My exam board syllabus doesn't contain graph sketching at all so I pretty much taught myself.:P I solved the c1 c2 c3 module textbook lessons for graphs to get the basics. I used to do this thing where my friend gave me a function,I used to sketch the graph and then check using a graph plotter. That's good for practice, you could do that.
I'm not that good at those either to be honest. My exam board syllabus doesn't contain graph sketching at all so I pretty much taught myself.:P I solved the c1 c2 c3 module textbook lessons for graphs to get the basics. I used to do this thing where my friend gave me a function,I used to sketch the graph and then check using a graph plotter. That's good for practice, you could do that.
Hey, I'm trying the Mathematics Interview Questions, was wondering if I could get a hint at differentiating x^x. I'm fairly sure you can't differentiate as normal, I'm tempted to differentiate in the same manner you differentiate exponential functions; however there are two variables. I can see that as x increases the function increases very quickly, similar to an exponential but far faster. Tip?
Also, just a question relating to the MAT. When marking the multiple choice, are any marks at all given for working?
the way I did it when I first saw the question was to rewrite x^x as e^ln(x^x) and then rewrite it as e^xln(x) which isn't too hard to differentiate using the product and chain rules.
q .15 1/(1-lnx) =-x/-x(1-lnx) take 1-lnx=t -1/x=dt also since 1-lnx=t.. 1-t=lnx thus e^(1-)t=x then youre left with the integral of -e^(1-t) which is e^(1-t) +constant replace t with 1-lnx you have e^lnx=x ANS x + constant
1/(1-lnx) =-x/-x(1-lnx) take 1-lnx=t -1/x=dt also since 1-lnx=t.. 1-t=lnx thus e^(1-)t=x then youre left with the integral of -e^(1-t) which is e^(1-t) +constant replace t with 1-lnx you have e^lnx=e ANS e + constant
I did the same thing but I made a stupid mistake with the value of x . Silly me Shouldn't it be -e + c though?
q .15 1/(1-lnx) =-x/-x(1-lnx) take 1-lnx=t -1/x=dt also since 1-lnx=t.. 1-t=lnx thus e^(1-)t=x then youre left with the integral of -e^(1-t) which is e^(1-t) +constant replace t with 1-lnx you have e^lnx=e ANS e + constant
q .15 1/(1-lnx) =-x/-x(1-lnx) take 1-lnx=t -1/x=dt also since 1-lnx=t.. 1-t=lnx thus e^(1-)t=x then youre left with the integral of -e^(1-t) which is e^(1-t) +constant replace t with 1-lnx you have e^lnx=e ANS e + constant
Your reasoning is a bit off, it looks like you missed the original u that you had substituted. It actually should be,
∫−x(1−lnx)−xdx=∫−xt1dt
Substituting for x=e^(1-t) thus gives,
∫t−1et−1du
Which is a non-elementary integral, as it simplifies to form with the exponential integral. Also on a related note, the second last-line is also incorrect as e^{ln x} = x.
q .15 1/(1-lnx) =-x/-x(1-lnx) take 1-lnx=t -1/x=dt also since 1-lnx=t.. 1-t=lnx thus e^(1-)t=x then youre left with the integral of -e^(1-t) which is e^(1-t) +constant replace t with 1-lnx you have e^lnx=e ANS e + constant
(e + constant)'=e I do not understand your method.....
Your reasoning is a bit off, it looks like you missed the original u that you had substituted. It actually should be,
∫−x(1−lnx)−xdx=∫−xt1dt
Substituting for x=e^(1-t) thus gives,
∫t−1et−1du
Which is a non-elementary integral, as it simplifies to form with the exponential integral. Also on a related note, the second last-line is also incorrect as e^{ln x} = x.
So when we are interviewed those type of questions, we do not have to give the final answer, right?
Your reasoning is a bit off, it looks like you missed the original u that you had substituted. It actually should be,
∫−x(1−lnx)−xdx=∫−xt1dt
Substituting for x=e^(1-t) thus gives,
∫t−1et−1du
Which is a non-elementary integral, as it simplifies to form with the exponential integral. Also on a related note, the second last-line is also incorrect as e^{ln x} = x.
derivative of 1-lnx = -1/x so answer is X + c
haha that was silly mistake ! spry . e^lnx=x my bad