If f(y) = e^(9y) + e^(-9y), find f'(y). Use exact values.

Mathematics

1 answer:

Westkost [7]3 years ago

5 0

Given:

$f(y)=e^{9y}+e^{-9y}$

To find:

The f'(y).

Solution:

Chain rule of differentiation:

$[f(g(x))]'=f'(g(x))g'(x)$

Differentiation of exponential:

$\dfrac{d}{dx}e^x=e^x$

We have,

$f(y)=e^{9y}+e^{-9y}$

Differentiate with respect to y.

$f'(y)=e^{9y}\dfrac{d}{dx}(9y)+e^{-9y}\dfrac{d}{dx}(-9y)$

$f'(y)=e^{9y}\cdot 9(1)+e^{-9y}\cdot (-9)(1)$

$f'(y)=9e^{9y}-9e^{-9y}$

Therefore, the differentiation of the given function is $f'(y)=9e^{9y}-9e^{-9y}$ .

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Find the coefficient of the fourth term of (x+2)^5

rjkz [21]

The coefficients of the binomial expansion $(a+b)^n$ , where n is the row number, is given in the Pascal's triangle shown below.

First, to find the coefficient of the fourth term of (x+2)^5 we look at row 5, term 4. The coefficient there is 10.

But, we must also remember that the term 2 also is taken to a certain power here. Mainly , for each term, the power of 2 is as follows:

2^0, 2^1, 2^2, 2^3=8.

So, in total we have: 10*8=80.

Second, to find the coefficient of the third term of (3x-1)^5 we again go to the row 5, this time term 3 and we have 10 there. Now we must check how each of (3x) and 1 expand, now being careful about the sign as well.

we have:
(3x)^5 (1) -(3x)^4 (1) (3x)^3(1)=27x^3.

Thus, the coefficient of the third term is 27*10=270.

Third, we want to find the coefficient of the third term of (a+5b^2)^4. We look at row 4, term 3. There we have 6.

The terms a and 5b^2 are as follows:

a^4 (5b^2)^0 a^3 (5b^2)^1 a^2 (5b^2)^2=25a^2b^4

Thus, the coefficient is 25*6=150.

Answer:

80; 270; 150