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Differentiation question

16 viewed last edited 1 year ago
Timothy Lee
1
I have two functions here: &nbsp; H(p) = 0,03p^(10) + 1,2p^(6) - p This would then be H'(p) = 0,3p^(9) + 7,2p^(5) - 1 And I think I understand this because -p is -1p &nbsp; But in this function: K(x) = 5x^(7) + ax^(2) + b The answer is K'(x) = 35x^(6) + 2ax &nbsp; I don't understand where the b is now and why a is in there, I think it may be because a is multiplying, but why does nothing happen with b? Wouldn't b be 1 then? Or does it need to be a letter which is in K(..) to count? (so x). And I also thought a would be 1 actually.
laxman
0
You are thinking in a wrong way. I recommend you to go with basics of the differentiation. Here a, b are the constants like 1,2,3,4......... And "p" is a variable like x ,y.... The differentiation of the constant is "0" \frac{d}{dx}(5) = 0 A Variable is a symbol for a number we don't know yet. It is usually a letter like x or y. A number on its own is called a Constant. A Coefficient is a number used to multiply a variable The defferentiation of a coefficient $\frac{d}{dx} 9x = 9 [ \frac{dx}{dx}]$ = 9*1 $\frac{d}{dx}[ 9x^2] = 9 [ \frac{dx^n}{dx}] = 9nx^{n-1}$ We have some formulas like this go througt them.
Sangeetha Pulapaka
0
The process of finding the derivative is called differentiation. The given function is H(p) = 0.03 p ^{10}+ 1.2 p{^6} - p When we differentiate once with respect to p, the power of the variable p is multiplied by the coefficient in front of it so the derivative of p ^{10} = 10 p{^9} In general the derivative of n^{th} term \frac{d}{dn}{a^n} = n a ^{n-1} \frac{dh}{dp} = 10 \times 0.03 {p}{^9}+ 6 \times1.2 {p }{^5} - \frac {dp}{dp} since \frac{dp}{dp}= 1 This can be written as H'(p) = 0.3 p^{9} + 7.2 p ^{5} - 1 K(x) = 5x ^{7} + a x^{2} + b Here we are differentiating with respect to x and using the same differential rules, k'(x) = 7 \times 5 x {^6} + 2xa + 0 since the differentiation of x^{2}= 2x and the differentiation of b with respect to x is 0. so k'(x) = 35 x{^6} +2 xa