queen salote college calculus revision on derivatives differentiation 2020: q3 a). find the derivative of…

Explanation:

Step1: Recall chain - rule for i

If $y = ae^{bx}$, then $y^\prime=abe^{bx}$. Here $a = 2$ and $b=-5$. $f^\prime(x)=2\times(- 5)e^{-5x}$

Step2: Recall product - rule for ii

The product - rule states that if $y = u(x)v(x)$, then $y^\prime=u^\prime(x)v(x)+u(x)v^\prime(x)$. Let $u(x)=e^{2x}$ and $v(x)=\cos(3x)$. First, find $u^\prime(x)$ and $v^\prime(x)$. For $u(x)=e^{2x}$, by the chain - rule $u^\prime(x)=2e^{2x}$. For $v(x)=\cos(3x)$, by the chain - rule $v^\prime(x)=-3\sin(3x)$. Then $f^\prime(x)=2e^{2x}\cos(3x)+e^{2x}\times(-3\sin(3x))$.

Answer:

i. $f^\prime(x)=-10e^{-5x}$ ii. $f^\prime(x)=2e^{2x}\cos(3x)-3e^{2x}\sin(3x)$