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You are right, never stop learning because it's the first time I have come across finding the derivative of an inverse function. I have always done the derivative of functions, but not their inverses. This is very informative.
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The formula for the derivative of an inverse function is nothing more than a peculiar application of the chain rule. Suppose f and g are inverse functions of one another. Then f(g(x)) = x. It follows that d[f(g(x))]/dx = f'(g(x))·g'(x) = dx/dx = 1. Hence g'(x) = 1/f'(g(x)). Considering your example, where we know f(x) = x^2 but we don't know that g(x) = sqrt[x], or if we do know that g(x) is sqrt[x], we don't know how to compute its derivative, we can apply the formula in the following way: We want g'(9). f'(x) = 2x => f'(g(x)) = 2g(x) => f'(g(9)) = 2g(9). Now, f(3) = 9 => g(f(3)) = 3 = g(9). Thus f'(g(9)) = 2·3 = 6. Finally, g'(9) = 1/f'(g(9)) = 1/6 ◼
Bro, you saved my exam results. Thank you so much
36 year old engineer here almost shed a tear seeing someone who looks like me teaching this
This guys energy is unmatched. I was watching this to study for my upcoming midterm and this made my entire day. Subscribed. I can't wait to come back here for studying
Thank you. I'm glad you find them entertaining.
I have a calculus test in the morning and with the way you just explained this, it has cut down hours of limbo the text books leave us in. Hats off 🧢.
This has been the most amazing, well thought out, constructed, well spoken analysis of any difficult math I have seen in my entire life, you are the reason I am going to do well on my test tomorrow.
Good luck!
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The concept of inverse and reciprocal had always been the same thing to me, but I've just learned they're not the same.
You're a great teacher sir!
Thank you so much..
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Beyond grateful to have stumbled upon this channel, not only is he super articulate with the way he explains problems, but he also manages to keep my attention and focus on what he is explaining. THANK YOUU
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You are right, never stop learning because it's the first time I have come across finding the derivative of an inverse function.
I have always done the derivative of functions, but not their inverses.
This is very informative.
Thank you for this feedback. And thank you for watching my videos. I appreciate you.
Thank you so much for this video! I've been trying to solve this concept for a while and you made it so simple and easy for me to understand. You are an amazing teacher!
I have looked through multiple calculus books and watched several different youtube vids and could not get this concept. This guy made it SO clear, and he did with energy and enthusiasm. THANK YOU!
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I'm glad it helped
Thanks for this video! Neither my textbook nor my calc professor could explain this in a way I could understand, but your explanation made it so clear.
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MAN your passion and energy makes me excited to learn maths.. it is true when they say if you have a great teacher you will truely love a subject...thank you🤗
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*EUROPEAN SPOTTED*
Such an amazing video on derivative of inverse function. Concept clear, no doubts. Thankyou from India 🙏
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I'm glad it helper
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Thank you. Glad you find my video helpful and credits to you still learning 😊
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EXCELLENT EXPLANATION!!
The formula for the derivative of an inverse function is nothing more than a peculiar application of the chain rule. Suppose f and g are inverse functions of one another. Then f(g(x)) = x. It follows that d[f(g(x))]/dx = f'(g(x))·g'(x) = dx/dx = 1. Hence g'(x) = 1/f'(g(x)).
Considering your example, where we know f(x) = x^2 but we don't know that g(x) = sqrt[x], or if we do know that g(x) is sqrt[x], we don't know how to compute its derivative, we can apply the formula in the following way: We want g'(9). f'(x) = 2x => f'(g(x)) = 2g(x) => f'(g(9)) = 2g(9). Now, f(3) = 9 => g(f(3)) = 3 = g(9). Thus f'(g(9)) = 2·3 = 6. Finally, g'(9) = 1/f'(g(9)) = 1/6 ◼
That is true. Good explanation.
Very good video