{"id":38297,"date":"2025-06-26T09:07:11","date_gmt":"2025-06-26T09:07:11","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=38297"},"modified":"2025-06-26T09:07:12","modified_gmt":"2025-06-26T09:07:12","slug":"given-y-e3t-find-the-following-derivatives-as-functions-of-t-dy-dt-3te3t-e3t-e-t-d2y-dt2-9t-6e-6t","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/given-y-e3t-find-the-following-derivatives-as-functions-of-t-dy-dt-3te3t-e3t-e-t-d2y-dt2-9t-6e-6t\/","title":{"rendered":"Given y = e^3t, find the following derivatives as functions of t: dy\/dt = (3te^(3t) + e^(3t)) – e^(-t) d^2y\/dt^2 = (9t + 6)e^(-6t)"},"content":{"rendered":"\n

Given y = e^3t, find the following derivatives as functions of t: dy\/dt = (3te^(3t) + e^(3t)) – e^(-t) d^2y\/dt^2 = (9t + 6)e^(-6t)<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

Let’s begin by carefully analyzing the function and computing the derivatives step by step.<\/p>\n\n\n\n

We are given:
y=e3ty = e^{3t}y=e3t<\/strong><\/p>\n\n\n\n

\n\n\n\n

First derivative: dydt\\frac{dy}{dt}dtdy\u200b<\/strong><\/h3>\n\n\n\n

To differentiate y=e3ty = e^{3t}y=e3t, apply the chain rule<\/strong>.
The derivative of eue^{u}eu with respect to ttt, where u=3tu = 3tu=3t, is:dydt=ddt(e3t)=3e3t\\frac{dy}{dt} = \\frac{d}{dt}(e^{3t}) = 3e^{3t}dtdy\u200b=dtd\u200b(e3t)=3e3t<\/p>\n\n\n\n

So the correct first derivative<\/strong> is:dydt=3e3t\\frac{dy}{dt} = 3e^{3t}dtdy\u200b=3e3t<\/p>\n\n\n\n

But your provided expression for the first derivative is:dydt=(3te3t+e3t)\u2212e\u2212t\\frac{dy}{dt} = (3t e^{3t} + e^{3t}) – e^{-t}dtdy\u200b=(3te3t+e3t)\u2212e\u2212t<\/p>\n\n\n\n

This does not match the derivative of e3te^{3t}e3t. It appears that the expression provided might come from a different function or a misunderstanding.<\/p>\n\n\n\n

\n\n\n\n

Second derivative: d2ydt2\\frac{d^2y}{dt^2}dt2d2y\u200b<\/strong><\/h3>\n\n\n\n

Now, differentiate dydt=3e3t\\frac{dy}{dt} = 3e^{3t}dtdy\u200b=3e3t again:d2ydt2=ddt(3e3t)=3\u22c53e3t=9e3t\\frac{d^2y}{dt^2} = \\frac{d}{dt}(3e^{3t}) = 3 \\cdot 3e^{3t} = 9e^{3t}dt2d2y\u200b=dtd\u200b(3e3t)=3\u22c53e3t=9e3t<\/p>\n\n\n\n

So the correct second derivative<\/strong> is:d2ydt2=9e3t\\frac{d^2y}{dt^2} = 9e^{3t}dt2d2y\u200b=9e3t<\/p>\n\n\n\n

But your given second derivative was:d2ydt2=(9t+6)e\u22126t\\frac{d^2y}{dt^2} = (9t + 6)e^{-6t}dt2d2y\u200b=(9t+6)e\u22126t<\/p>\n\n\n\n

Again, this does not correspond to the second derivative of e3te^{3t}e3t. That expression likely came from a different function entirely, possibly involving products or more complex composition.<\/p>\n\n\n\n

\n\n\n\n

Conclusion<\/strong><\/h3>\n\n\n\n

For the function y=e3ty = e^{3t}y=e3t, the correct derivatives as functions of ttt are:<\/p>\n\n\n\n

    \n
  • First derivative<\/strong>: dydt=3e3t\\frac{dy}{dt} = 3e^{3t}dtdy\u200b=3e3t<\/li>\n\n\n\n
  • Second derivative<\/strong>: d2ydt2=9e3t\\frac{d^2y}{dt^2} = 9e^{3t}dt2d2y\u200b=9e3t<\/li>\n<\/ul>\n\n\n\n

    These results follow directly from applying basic rules of differentiation for exponential functions. Exponential functions like ekte^{kt}ekt are particularly straightforward to differentiate, as their form is preserved and only multiplied by the constant from the exponent.<\/p>\n\n\n\n

    \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

    Given y = e^3t, find the following derivatives as functions of t: dy\/dt = (3te^(3t) + e^(3t)) – e^(-t) d^2y\/dt^2 = (9t + 6)e^(-6t) The Correct Answer and Explanation is: Let’s begin by carefully analyzing the function and computing the derivatives step by step. We are given:y=e3ty = e^{3t}y=e3t First derivative: dydt\\frac{dy}{dt}dtdy\u200b To differentiate y=e3ty […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-38297","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/38297","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=38297"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/38297\/revisions"}],"predecessor-version":[{"id":38301,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/38297\/revisions\/38301"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=38297"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=38297"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=38297"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}