![]() It is a clever use of the fact that adding and subtracting the same value does not change an expression. Also watching Implicit differentiation and logarithmic differentiation has led me to believe this could be proven with natural logarithm but how? That way I can always a problem down into what I know and therefore I can always find the formula for whatever problem I might have. That is, it's easier for me to understand something instead memorizing the entire thing. ![]() So replace F(x) and F(x+h) with corresponding terms and you get what Sal got, did I get this right? I really do need to get this right otherwise my brain will refuse to memorize it. Thankfully we already know what the product rule is so we sort of know what we need to add the expression but there has to be a more intuitive way to prove the product rule? Also he's multiplying the entire derivations of 2 functions so why isn't it ((f(x+h)-f(x))*(g(x+h)-g(x)))/h ? Am guessing because the product of 2 derived functions is not the same as derivation of products of functions? This would mean that f(x)*g(x) is now one new function, it's not 2 seperate functions but one as in multiplications where ab is not 2 terms but 1 term, so f(x)*g(x)=F(x), if we evaluate F(x) we get (F(x+h)-F(x))/h and like I said, F(x)=f(x)*g(x). Therefore, the only solution is the pointThe little trick he uses is not sufficient enough for me to prove to myself the product rule. The derivative of f isīut x=0 cannot be a solution since. Thus, we need to find all values x in the domain of f which satisfy f'( x)=-1. The slope of the given line y = 12- x is y'=-1. SOLUTION 21 : Find all points ( x, y) on the graph of with tangent lines parallel to the line y + x = 12. Thus, the slope of the line perpendicular to the graph at is Thus, the slope of the line tangent to the graph at is The graph of f will be perpendicular to the line tangent to the graph of f. SOLUTION 20 : Find an equation of the line perpendicular to the graph of The slope of the line tangent to the graph at is The slope of the tangent line follows from the derivative of y. So that the tangent line passes through the point SOLUTION 19 : Find an equation of the line tangent to the graph of Thus, the values of x which solve f'( x)=0 are )īut is never zero since exponentials are always positive. (Now factor out the common terms of, and. For what values of x is f'( x)=0 ? Begin by differentiating f using the triple product rule. Differentiate y using the triple product rule. Then add the three new products together. Each time differentiate a different function in the product. Here is an easy way to remember the triple product rule. Group functions f and g and apply the ordinary product rule twice. Thenīut x = 0 is not in the domain of function f since is not defined. For what values of x is f'( x)=0 ? Begin by differentiating the function using the product rule. Thus, the only values of x which solve f'( x) = 0 are ![]() Thenīut can never be zero since an exponential is always positive. ![]() Thus, the values of x which solve f'( x) = 0 areĬlick HERE to return to the list of problems. For what values of x is f'( x) = 0 ? Begin by differentiating the function using the product rule. SOLUTIONS TO DIFFFERENTIATION OF FUNCTIONS USING THE PRODUCT RULE ![]()
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