fg<\/em> is also differentiable, and<\/p>\n<\/p>\n
here is another way to state the rule in a more compact way:<\/p>\n
<\/p>\n
many textbooks write the terms in different orders. here is yet another version of the rule, this time using leibniz notation<\/em>.<\/p>\n<\/p>\n
example 1<\/h3>\n
find the derivative of f<\/em>(x<\/em>) = x<\/em>5<\/sup> sin x<\/em>.<\/p>\nthis function is the product of two functions, u<\/em> = x<\/em>5<\/sup> and v<\/em> = sin x<\/em>. using the product rule, <\/p>\n<\/p>\n
note, the correct derivative is not<\/strong> found by just taking the product of the derivatives. in other words, f<\/em> '(x<\/em>) ≠ 5x<\/em>4<\/sup> cos x<\/em>.<\/p>\nexample 2<\/h3>\n
in the next example, we’ll see that the product rule can even be used when very little information is known about the individual functions.<\/p>\n
suppose f<\/em> and g<\/em> are differentiable at x<\/em> = 4, and suppose we know that f<\/em>(4) = -3, g<\/em>(4) = 2, f<\/em> '(4) = 7, and g<\/em> '(4) = 3. let h<\/em> = fg<\/em>. what is the value of h<\/em> '(4)?<\/p>\nbe careful!<\/em> the answer is not<\/strong> 21. instead, set up the rule and then plug in the known information:<\/p>\n<\/p>\n
practice problems<\/h2>\n
now’s your chance to practice using the product rule! solutions are provided at the end.<\/p>\n
\n- find the derivative of .<\/li>\n
- if y<\/em> = sec x<\/em>, what is y<\/em> ''?<\/li>\n
- (requires chain rule)<\/em> differentiate g<\/em>(t<\/em>) = cos(t<\/em> sin t<\/em>).<\/li>\n
- find the equation of the tangent line to the curve at x<\/em> = 1.<\/li>\n<\/ol>\n
<\/p>\n
solutions<\/h3>\n\n- <\/li>\n
- the first derivative is: y<\/em> ' = sec x<\/em> tan x<\/em>. the product rule is required to find the second derivative:\n
<\/li>\n
- use the chain rule first, and then product rule for the derivative of the “inside” function.
\n\n<\/li>\n - to find the equation of the tangent line, we will need the slope. and that requires finding a derivative.\n
<\/p>\n
plug in x<\/em> = 1 to find the slope.<\/p>\n<\/p>\n
finally, we need the y<\/em>-coordinate of the point. plug x<\/em> = 1 back into the original function to get: y<\/em> = e<\/em>. now we can put together the tangent line equation:<\/p>\n\n<\/ol>\n","protected":false},"excerpt":{"rendered":"
the product rule is just one of many essential derivative rules. together with the sum\/difference rule, power rule, quotient rule, and chain rule, these rules form the backbone of our methods for finding derivatives. in this article i’ll explain what the product rule is and how to use it in typical problems on the ap […]<\/p>\n","protected":false},"author":223,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[240],"tags":[241],"ppma_author":[24932],"acf":[],"yoast_head":"\n
ap calculus review: product rule - magoosh blog | high school<\/title>\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\t\n