in this video we're going to talk about how to find the derivative using the definition of the derivative formula so basically we need to find the derivative of a function using the limit process and f prime of x represents the derivative of f of x and it's equal to the limit as h approaches zero of f of x plus h minus f of x divided by h so that's the formula that we need to use so let's say if f of x is a linear function 5 x minus 4. what is f prime of x what's the first derivative of this function so go ahead and try that now what is f of x plus h how can we find that well if f of x is five x minus four then f of x plus h is going to be five times x plus h minus four so all you need to do is in f of x wherever you see an x replace it with x plus h so now let's plug in everything into this formula so it's going to be the limit as h approaches 0 and then it's going to be 5 x plus h minus 4 and then minus f of x which f of x is 5 x minus 4. and so we're going to divide all of this by h now let's simplify so first we need to distribute 5 to x plus h and don't forget to rewrite the limit expression until you get your final answer so it's going to be 5x plus 5h minus 4 and then distribute the negative sign so it's going to be negative 5x plus 4 divided by h now five x and negative five x they add up to zero negative four plus four is equal to zero so we're left over with the limit as h approaches zero 5 h divided by h and the h terms cancel h divided by h is 1 and so the limit as h approaches 0 for a constant like 5 since there's no h in this expression anymore this is simply equal to five so that's the derivative of five x minus four it's five now let's try another example so let's say if f of x is equal to x squared what is the first derivative of the function go ahead and use the definition of the derivative to find f prime of x so first let's use the formula f prime of x is equal to the limit as h approaches zero of f of x plus h minus f of x divided by h so if f of x is equal to x squared what is f of x plus h that's the first thing you need to decide so all we have to do is replace x with x plus h so it's going to be x plus h squared so we have the limit as h approaches 0 x plus h squared minus f of x which is originally just x squared divided by h now we need to foil x plus h squared so i'm going to expand it for now so what we really have is x plus h times another x plus h so let's foil x times x that's going to be x squared and since i'm running out of space i'm going to put this on the top and then x times h that's xh and then we have to multiply h times x which is also xh and finally h times h that's going to be h squared and then minus x squared all divided by h now let's combine like terms and let's cancel x squared and negative x squared adds up to zero and xh plus xh well that's 2xh and so we're going to have the limit as h approaches 0 for 2xh plus h squared divided by h now what do you think we need to do at this point once you get to this step what we need to do is factor out an h if we take out an h the gcf 2xh divided by h is 2x and h squared divided by h well that's h now we could cancel the h variables on the outside h divided by h is one so this gives us the limit as h approaches zero for two x plus h so now we're gonna apply this limit expression we're going to replace h with 0. so it's going to be 2x plus 0 which is 2x so if f of x is equal to x squared the derivative f prime of x is equal to 2x based on the definition formula of the derivative and that's the answer now let's work on another example so let's say if we have one over x what is the first derivative of one over x by the way feel free to pause the video and try these examples yourself so first let's find f of x plus h so if we replace x with x plus h we're going to get 1 over x plus h now let's start with the formula f prime of x is equal to the limit as h approaches zero f of x plus h minus f of x divided by h now f of x plus h we said it's one over x plus h and f of x is one over x and this is all divided by h so here we have a complex fraction what should we do in a situation like this if you have a complex fraction multiply the top and the bottom by the common denominator of these two fractions so the common denominator is going to be x times x plus h whatever you do to the top you must also do to the bottom of the complex fraction now we need to multiply so if we take this fraction and multiply by x times x plus h the x plus h terms will cancel eliminating this fraction and what we have left over is simply x so right now we're going to have the limit as h approaches 0 and on top we have an x right now now let's take the second fraction multiply by that term so if we do that x will cancel and we're going to get x plus h with a negative sign in front of it so this is going to be negative x plus h and on the bottom we simply just need to write these things together so it's going to be h times x times x plus h now let's distribute the negative sign so we're going to have the limit as h approaches 0 x minus x minus h x plus negative x adds up to zero and so this is going to give us the limit as h approaches zero negative h divided by h times x times x plus h negative h divided by h is negative one so now we have the limit as h approaches zero negative one over x times x plus h as soon as you cancel that h you can now use direct substitution so we can apply this limit expression let's replace h with zero now x plus zero is x so we have x times x which is x squared so the answer is negative one divided by x squared and that is the derivative of one over x now let's say that f of x is equal to the square root of x what is the first derivative f prime of x go ahead and try that so let's start with the formula f prime of x is equal to the limit as h approaches zero f of x plus h minus f of x over h now in this problem what do you think f of x plus h represents now don't forget all we need to do is replace x with x plus h so it's going to equal the square root of x plus h and f of x is simply the square root of x now what should we do if we have radicals in a fraction to simplify this expression you need to multiply the top and the bottom by the conjugate of the numerator so the conjugate is going to have the same expression square root x plus h and square root x the difference is instead of a negative sign we are going to have a positive sign so now let's foil what we have on top so the square root of x plus h times itself the square roots will cancel and you're going to get x plus h and then we have these two terms which is going to be plus square root x square root x plus h and then if we multiply those two terms negative square root x times square root x plus h and finally the square root of x times the square root of x is simply x and don't forget about the negative sign and on the bottom don't distribute simply just i would recommend just rewriting it now the square root of x times the square root of x plus h we have a positive term and a negative one so they're going to add up and cancel to zero x to negative x will also add up to zero so now we're left with the limit as h approaches zero and it's h divided by h times the square root of x plus h plus the square root of x and as you know h divided by h is one so we're left with the limit as h approaches zero and we have this expression now what we need to do is direct substitution we need to replace h with zero at this point so x plus zero will simply be x now the square root of x plus the square root of x there's a coefficient of one one plus one is two so you should get one over 2 square root x by the way this limit expression should no longer be here when you replace h with 0 this expression disappears so it shouldn't be in this step anymore this is the final answer so that's the first derivative of the square root of x now what if the square root is in the bottom of a fraction like if we have 8 over the square root of x what's the first derivative of this function well first let's find f of x plus h so it's going to equal that now let's use the limit process to find the derivative so f of x plus h it's going to be 8 over the square root of x plus h and i'm forgetting the limit expression this is why it's always good to rewrite the formula just to avoid making mistakes so we have the limit as h approaches zero and then f of x plus h is eight over the square root of x plus h and then minus f of x and then divided by h so now we have a complex fraction with radicals so basically we have two problems combined into one we have the one over x problem with the square root of x problem so what i recommend is eliminating the fractions within the larger fractions so i'm going to multiply by the common denominator just to begin so it's going to be the square root of x times the square root of x plus h so these terms will cancel leaving behind eight times the square root of x and then when i multiply this fraction by that term the square root of x will cancel leaving behind these two so it's going to be negative 8 square root x plus h and on the bottom simply rewrite these things together so it's going to be h times the square root of x times the square root of x plus h now what should we do at this point what's our next move here so since we have radicals in the numerator we need to multiply the top and the bottom by the conjugate of the numerator and so that's going to be eight square root x plus a square root x plus h divided by the same thing so let's foil this stuff on top so 8 square root x times itself what is that going to give us well we know 8 times 8 is 64. and the square root of x times the square root of x is x next we have 8 times 8 which is 64 and then square root x square root x plus h and then we have the two middle terms this is going to be negative 64 square root x square root x plus h and then finally negative eight times eight and then square root x plus h times itself which is just going to be x plus h and on the bottom just rewrite everything now this is a very long problem but if you have a problem like this on a test this is what you have to do so first we could cancel these two terms and then let's distribute negative 64 to x plus h so what we have now is 64x and then negative 64 x and negative 64 h divided by everything on the bottom now let's cancel those two terms and so we're left with limit as h approaches 0 negative 64 h divided by everything that is in the denominator which is a lot of stuff now in the next step we could cancel h because as soon as we can do that we can use direct substitution we can get rid of the other h variables in the bottom right now we have the limit as h approaches 0 negative 64 over square root x square root x plus h and so forth so now i'm going to replace h with 0 so i'm no longer going to write the limit expression so it's negative 64. and then we have square root x now this is going to be square root x plus 0 which is also square root x and then 8 times the square root of x and then we're going to replace h of 0 there so that's going to be 8 square root x as well now the square root of x times the square root of x is x and then 8 plus 8 is 16 and then we could divide negative 64 by 16. so we're going to get negative 4 over x square root x and so you can leave the answer like that if you want to x square root x is also x to the three halves you could write it that way too but that's the answer now the last example i'm going to go over is a polynomial function so let's say we have f of x is equal to x squared minus five x plus nine what's the first derivative of this function so first what's f of x plus h let's decide that so instead of writing x squared we're going to write x plus h squared instead of writing 5x is going to be 5 times x plus h so then f prime of x is going to be the limit as h approaches zero and then f of x plus h that's x plus h squared minus five x plus h plus nine and then minus f of x which is x squared minus five x plus nine don't forget to distribute the negative sign and this is all divided by h so first we need to expand x plus h squared already now we've done that early in this video so x plus h times x plus h we found that it was x squared plus two x h plus h squared now we need to distribute the five so it's going to be negative 5x minus 5h plus 9 and then distribute this negative sign so it's going to be negative x squared plus 5x minus 9 all divided by h so x squared and negative x squared can be cancelled negative 5x and 5x will disappear and 9 and negative 9 we can get rid of so now we have left over the limit as h approaches zero so we have h squared plus 2xh minus 5h divided by h now let's factor out the greatest common factor in the numerator which is h so h squared divided by h is h 2xh divided by h is 2x negative 5h divided by h is negative 5. so now we could cancel those two and so we have the limit as h approaches zero h plus two x minus five now let's use direct substitution let's replace h with zero so zero plus two x minus five is simply two x minus five so that's the first derivative of x squared minus five x plus nine it's two x minus five you