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Solve the given problems by finding the appropriate derivative.The resistance $R$ (in $\Omega$ ) of a certain wire as a function of the temperature $T$ (in $^{\circ} \mathrm{C}$ ) is given by $R=16.0+0.450 T+0.0125 T^{2}$ Find the instantaneous rate of change of $R$ with respect to $T$ when $T=115^{\circ} \mathrm{C}$

Calculus 1 / AB

Chapter 23

The Derivative

Section 5

Derivatives of Polynomials

Derivatives

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Welcome to new Madrid. In the current problem. We are given the resistance are in terms which is uh units are in omega. And this is given to be 16 plus 0.450 T glass. zero point you want to find the square where T. Is the temperature separator Bill believe in two years. Okay. So do you want to know the instantaneous rate of change? You are So in stan Taney. Isse change of art week, respectable then temperate it T. That is D R D. T. We want. So that would be DDT off 16 plus 0.450 T Plus 0.0125 T sq. Get it. So we will have now we'll split the terms individually with respect to the derivative and write the expression 0.450 into DD. T. of T. Plus 0.0125 into the D. T. of T. Square. Therefore we will have zero plus 0.450 into one plus 0.125 in. Due to T. It's zero point 0.45. We do plus This would be zero five door set up. Keep okay so therefore At musical 215. So we are sweetie When T. is equals to 115 0.45 plus 115 Into 0.0250. That would be let us try didn't despite first. And so we will have little point three point 3 to 5. We have cheap blanks. She 25. Which if we Want we can simplify to be 3.3. Now what will be the unit? This has unit omega courage and T. Has unit The recent. Yes so this will be 3.30mega par degree C. So I hope you could understand let me know if you have any questions

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