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... The 2nd indeterminate limit can also be solved as follows ... in this case rewriting the numerator x - 5 as (4 + x) - 9 , and considering this expression as a difference of 2 squares ... (4+ x) - 9 = (SQRT(4 + x) - 3)(SQRT(4 + x) + 3) ... then cancelling common factor SQRT(4 + x) - 3 between top and bottom, giving us a solvable limit form in return ... lim(x->5)[SQRT(4 + x) + 3] = 6 ... solving this limit by factoring and cancelling ... thank you for your clear and instructive presentations ... take good care, Jan-W

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... Good day to you, Another way of solving your indeterminate limit, is to rewrite the denominator x in its original form as follows ... x = (4 + x) - 4 , and then treating this expression as a difference of 2 squares ... (SQRT(4 + x) - 2)(SQRT(4 + x) + 2) to finally cancelling the common factor (SQRT(4 + x) - 2) between numerator and denominator, to be left with a solvable limit form ... lim(x->0)[1/(SQRT(4 + x) + 2) = 1/4 ... solving by factoring I think I would call this ... thanks for sharing your instructive math channel with us, the interested viewers ... best wishes, Jan-W

Excellent explanation

Very great work, perfect explanation

Don't know how much you have to pay for AI voice, but you should consider switching even if you have to pay a bit more. Get an AI voice that knows a little math. "ln" is not "lawn" as in "mow the lawn". Once or twice might be charming but the whole video is about ln so "lawn" gets quickly annoying. Mike

God bless you

Thank you

You don’t need to say it’s lnx times 1 lol just let dv = dx

In the case of a pure inverse function, the integration by parts formula can be derived from the graph: int x dy=x*y-int y dx

1:07 Why all that talk 🤣 we know that eⁿ = x remplace it

Use first principles.

You kids get off my lawnx!

Can this be a method to prove the f'?

Hard matn😅

but if we look at x^3, wouldn't y= x+1 be a tangent to x^3 at x = 3 be x+24? The slope of x + 24 is 1 but with differentiation, slope of tangent = 2*3^2 = 2*9 = 18?

Thanks for making me understand more about this topic. I would like to ask does chain rule applied to all implicit equations? And if not, in what situation we can use it?

"lawn y"

Can we take the derivative using power rule by applying dy/dx=sin x *sin x?

Very comprehensive explanation

to the point explanation very great.

Quotient rule...

Also from cos^2(x)+sin^2(x)/cos^2(x) we get 1+tan^2(x) which is by identity of sec^2(x)-tan^(x)=1, we have sec^2(x).

Amazing , just keep on posting more stuff

Can this be done by implicit differentiation?

Tq sir ❤...

Subscribed to your channel and like 👍 your explanation to sir.. tq sir ❤ love you Teacher.

Sir tangent slope finding question please

Love you sir ❤ thanks a lot...

Thanks for the video 😊

Tabular method?

I memories that sin^2 equals to 1/2(1-cox2x) and just do it from there

Very useful tutorial.Thank you

thanks bro

Chain rule

Easy, it's x.x^(x-1), done.

Isn't it ILATE instead of LIATE

I never use this method. I use the D-I method. Much, much easier. And much easier to learn.

We can use the limit definition to...

J=-ln(cosx)+k

To simplify one can for final answer: 1/xln(lnx^2)

Sir! Please, explain once and for all the difference between logx and lnx. I get so many confused pieces of information and even on YT you see mathematicians saying that logx is 1/x. Also, I hear the word e for exponential in some videos where I try to explain that e is a number which can be used as a base for a logarithm but by itself it is not any exponential (or is it?) and it's called Euler's number. Thank you.

Great

Very nice, I've learned something new! Thank You!

The easiest way is to just use the definition of tanx as the ratio of sine to cosine. U-sub cosine and then that gives du=-sinxdx. Then you get -ln|cosx| or ln|secx|+C.

Not the simplified form, but you can pull out the factor of 4 as 199(4)(x+1)(2x^2+4x-3)^198=796(x+1)(2x^2+4x-3).

The third method is multiply sec(x)/sec(x). The integral should be sec(x)tan(x)/sec(x)dx. If you let u=sec(x) and du=sec(x)tan(x), the integral is du/u. The answer to this integral is ln(|u|)+C=ln(|sec(x)|)+C. If you want in terms of cos(x), then the answer is -ln(|cos(x)|)+C.

ur so underated omg

Help me in his question Cos^3xdx by using integration by party

J=xlnx-x+k