anonymous
  • anonymous
Suppose h(t) = t^2+14t+7. Find the instantaneous rate of change h(t) with respect to t at t=2.
Mathematics
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
Please show your work so I can see what you did. Thanks.
Xishem
  • Xishem
First differentiate with respect to t. This finds an equation for the rate of change: h'(t) = 2t + 14 Then, simply substitute 2 into the equation to get: h'(2) = 2(2) + 14 = 18
anonymous
  • anonymous
could you explain the differentiating part for me please

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
i understand it somewhat but would appreciate a clear explanation
anonymous
  • anonymous
?
Xishem
  • Xishem
When you take the derivative of a function, you follow a set of rules. The most basic rule you need to know is when you have just a constant in front of a variable to some power n. The general form for this type of differentiation is: \[f(x) = x^n\]\[f'(x) = nx^{n-1}\] In the case of this question, you can differentiate each term separately: \[f(x)=t^2+14t+7\]\[f'(x)=2t^1+14t^0+0 = 2t+14\] Whenever a constant is differentiated, it becomes 0. Does that make more sense now?
anonymous
  • anonymous
yes, THANKS!!!
Xishem
  • Xishem
Keep in mind that the power rule only works when you have a constant in front of the variable you differentiating in respect to. If you have products or quotients of expressions, it doesn't work correctly.
anonymous
  • anonymous
ok

Looking for something else?

Not the answer you are looking for? Search for more explanations.