Verified answer

Qu 1

f ( x ) = y = 7 e^(x cos x)

Let u = x cos x

du/dx = cos x - x sin x

y = 7 e^u

dy/du = 7e^u

dy/dx = 7e^u ( cos x - x sin x )

f ` ( x ) = dy/dx = 7e^(x cos x) ( cos x - x sin x )

Qu 2

f ( x )= [ 3x^2 + 3x + 7 ]^(1/2)

f ` ( x ) = (1/2) [ 3x^2 + 3x + 7 ]^(- 1/2) [ 6 x + 3 ]

f ` ( x ) is then : -

3 ( 2 x + 1 )

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2 [ 3x^2 + 3x + 7 ]^(1/2)

For the first problem, you must use the chain rule in order to find the derivative. If you remember, g(f(x)) is g'(f(x))*f'(x). In other words, you find the derivative of the outside like normal, but you multiply by the derivative of the inside. In this case, we have 7*e^(x*cos(x)). The derivative of x*cos(x) is, by the product rule, -x*sin(x) + cos(x). This gives us a final derivative of 7*e^(x*cos(x))*(-x*sin(x) + cos(x)).

For the second part, you also have to use the chain rule. Bring down the power and reduce it by one, then multiply by the derivative:

(1/2)*(3x^2 + 3x + 7)^(-1/2) * (6x + 3)

To get f'(3), all you must now do is plug in 3 for x, which gives us about 1.60123499.

f(x) = 7 * e^(x*cosx)

f'(x) = 7 * e^(x*cosx) * (cosx - x * sinx)

f(x) = (3x^2 + 3x + 7)^(1/2)

f'(x) = 1/2 * ((3x^2 + 3x + 7)^(-1/2) * (6x + 3)

f'(3) = 1/2 * ((3 * 3^2 + 3 * 3 + 7)^(-1/2) * (6 * 3 + 3)

f'(3) = 1/2 * (27 + 9 + 7)^(-1/2) * (18 + 3)

f'(3) = 21 / (2 * sqrt(43))

f'(3) = 1.6

You seem on the final music yet per possibility in basic terms getting slightly perplexed with the derivatives? it may be extra hassle-free to interrupt the g(x) into products and do the differentiation bit by using bit. as an occasion, enable g(x) = w(x)sin(x) + v(x)cos(x) the place w(x) = ax^2 + bx + c and v(x) = dx^2 + ex + f. Then by using the product rule we've g'(x) = w(x)cos(x) + w'(x)sin(x) -v(x)sin(x) + v'(x)cos(x) = [w(x) + v'(x)]cos(x) + [w'(x) - v(x)]sin(x). comparing this with the given expression for g'(x) shows we could have w(x) + v'(x) = 0 and w'(x) - v(x) = x^2, or ax^2 + bx +c + 2dx + e = 0 and 2ax + b - {dx^2 + ex +f} = x^2. comparing powers of x in those 2 equations capacity we could pick a = 0, b+ 2nd = 0 and c + e = 0 from the 1st equation and d = -a million, 2a - e = 0 and b - f = 0 from the 2nd equation. those are six equations for the six unkowns a,b, c, d, e and f and that i'm useful you are able to sparkling up from right here!