Math 1211 Written Assignment Unit 5

1. Find the derivative for the function Given: f(x) = 2 ex − 8 x d/dx (2 ex ) = 2 ⋅ d/dx e x = 2e x d/dx 8 x = 8 x ⋅ ln(8) Therefore, the derivative of the function f(x)=2 ex −8 x is: = 2 ex − 8 x ⋅ ln(8) 2. Find the derivative for the function Differentiating z 5 with respect to z using the power rule (d/dz z n = n ⋅ z n−1 ): d/dz (z 5 ) = 5z 5 – 1 = 5z 4 u = d/dz ′ e z = e z (since the derivative of e z is e z ) v = d/dz ln(z) = 1/z (using the derivative of ln(z)) ′ Applying the product rule: d/dz (e z l n(z)) = e z ⋅ ln(z) + e z ⋅ 1/z Therefore, the derivative of the function 𝑓 ’( 𝑧 ) = 5 𝑧 4 − 𝑒 𝑧 /z − 𝑒 𝑧 ln ( 𝑧) 3. Find the tangent line to at f (0) = 7 ′ 0 ln(7) + 4e 0 f (0) = ln(7) + 4 ′ f(0) = 7 0 + 4e 0 f(0) = 1 + 4 = 5 At x = 0 y=5. y−5 = (ln(7)+4) (x−0) y = 5 + (ln(7) + 4)x

4. Determine if is increasing or decreasing at the following points. (a) z=1 f (z) = d/dz ′ (z − 6) = 1 g (z) = d/dz ln(z) = 1/z ′ Now applying the product rule: G (z) = f (z)g(z) + f(z)g (z) ′ ′ ′ G (z) = 1 ′ ⋅ ln(z) + (z−6) ⋅ 1/z G (z) = ln(z) + z ′ – 6 / z To determine the behavior of G(z) at z=1, let's evaluate G (1): ′ G (1) = ln(1) + 1 ′ − 6 / 1= 0 − 5= −5 Since G (1) = −5 < 0, the function G(z) = (z−6)ln(z) is decreasing at z=1. ′ (b) z=5 G (5) = ln(5) + 5 – 6 / 5 ′ G (5) = ln(5) ′ – 1/5 Since ln(5) is positive and 1/5 is positive as well, G (5) = ln(5) ′ − 1/5 is positive overall. Therefore, as G (5)>0, the function G(z) = (z−6)ln(z) is increasing at z=5. ′ (c) z=20 The natural logarithm of 20 is positive and 7/10 is positive as well, hence G′(20) = ln(20) + 7/10 is positive overall. Therefore, as G′(20)>0, the function G(z) = (z−6)ln(z) is increasing at z=20. 5. Find the derivative for the function f(x) = (x + 1) x d/dx (ln(y)) = d/dx (x ⋅ ln(x+1)) 1/y ⋅ dy/dx = ln(x+1) + x ⋅ 1 / x + 1 1/(x+1) x ⋅ dy/dx = ln(x + 1) + x / x+1x = (x + 1) x ⋅ (ln(x + 1) + x / x+1) 6. Find the derivative for the function