Answer<\/b> ##4\/(16x^2 + 1)##<\/p>\n
Explanation<\/b> First recall that ##d\/dx[arctan x] = 1\/(x^2 + 1)##.<\/p>\n
Via the :<\/p>\n
1.) ##d\/dx[arctan 4x] = 4\/((4x)^2 + 1)##<\/p>\n
2.) ##d\/dx[arctan 4x] = 4\/(16x^2 + 1)##<\/p>\n
If it isn’t clear why ##d\/dx[arctan x] = 1\/(x^2 + 1)##, continue reading, as I’ll walk through proving the identity.<\/p>\n
We will begin simply with<\/p>\n
1.) ##y = arctan x##.<\/p>\n
From this it is implied that<\/p>\n
2.) ##tan y = x##.<\/p>\n
Using implicit differentiation, taking care to use the chain rule on ##tan y##, we arrive at:<\/p>\n
3.) ##sec^2 y dy\/dx = 1##<\/p>\n
Solving for ##dy\/dx## gives us:<\/p>\n
4.) ##dy\/dx = 1\/(sec^2 y)##<\/p>\n
Which further simplifies to:<\/p>\n
5.) ##dy\/dx = cos^2 y##<\/p>\n
Next, a substitution using our initial equation will give us:<\/p>\n
6.) ##dy\/dx = cos^2(arctan x)##<\/p>\n
This might not look too helpful, but there is a trigonometric identity that can help us.<\/p>\n
Recall ##tan^2alpha + 1 = sec^2alpha##. This looks very similar to what we have in step 6. In fact, if we replace ##alpha## with ##arctan x##, and rewrite the ##sec## in terms of ##cos## then we obtain something pretty useful:<\/p>\n
##tan^2(arctan x) + 1 = 1\/(cos^2(arctan x))##<\/p>\n
This simplifies to:<\/p>\n
##x^2 + 1 = 1\/(cos^2(arctan x))##<\/p>\n
Now, simply multiply a few things around, and we get:<\/p>\n
##1\/(x^2 + 1) = cos^2(arctan x)##<\/p>\n
Beautiful. Now we can simply substitute into the equation we have in step 6:<\/p>\n
7.) ##dy\/dx = 1\/(x^2 + 1)##<\/p>\n
And voil\u00e0 – there’s our identity.<\/p>\n","protected":false},"excerpt":{"rendered":"
Answer ##4\/(16x^2 + 1)## Explanation First recall that ##d\/dx[arctan x] = 1\/(x^2 + 1)##. Via the : 1.) ##d\/dx[arctan 4x] = 4\/((4x)^2 + 1)## 2.) ##d\/dx[arctan 4x] = 4\/(16x^2 + 1)## If it isn’t clear why ##d\/dx[arctan x] = 1\/(x^2 Read More …<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[],"tags":[],"class_list":["post-170604","post","type-post","status-publish","format-standard","hentry"],"_links":{"self":[{"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/posts\/170604","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/comments?post=170604"}],"version-history":[{"count":0,"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/posts\/170604\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/media?parent=170604"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/categories?post=170604"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/tags?post=170604"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}