## - Differentiation of the Exponential Function

$$y=e^x\longrightarrow y'=e^x$$

$$y=e^u(x) \longrightarrow y'=u' e^u(x)$$

LESSON 1
Differentiate the following
(a) $$y=e^{2x-3}$$
(b) $$y=2e^{\sin ⁡3x}$$
(c) $$y=xe^{x^2}$$
(d) $$\displaystyle y={e^{-2x}\over x+1}$$
(e) $$y=(1-e^{5x})^3$$
SOLUTION
(a) $$y=e^{2x-3}$$
$$u(x)=2x-3 → u'(x)=2$$
$$\displaystyle {dy\over dx}=2e^{2x-3}$$

(b) $$y=2e^{\sin ⁡3x}$$
$$u(x)=\sin ⁡3x → 3 \cos ⁡3x$$
$$\displaystyle {dy\over dx}=2[3\cos⁡ 3x] e^{\sin ⁡3x}$$
$$=6e^{\sin ⁡3x} \cos ⁡3x$$

(c) $$y=xe^{x^2}$$
• Use the PRODUCT RULE
$$\displaystyle {dy\over dx}=1e^{x^2}+x[2xe^{x^2}]$$
$$=e^{x^2}+2x^2e^{x^2}$$
$$=e^{x^2}[1+2x^2]$$

(d) $$\displaystyle y={e^{-2x}\over x+1}$$
• Use the QUOTIENT RULE
$$\displaystyle {dy\over dx}={-2e^{-2x}(x+1)-e^{-2x}(1)\over (x+1)^2}$$
$$\displaystyle ={-2xe^{-2x}-2e^{-2x}-e^{-2x}\over (x+1)^2}$$
$$\displaystyle ={-e^{-2x}(2x+2+1)\over (x+1)^2}$$
$$\displaystyle ={-e^{-2x} (2x+3)\over (x+1)^2}$$

(e) $$y=(1-e^{5x})^3$$
• Use the CHAIN RULE
$$\displaystyle {dy\over dx}=3(1-e^{5x})^2(-5e^{5x})$$
$$=-15(1-e^{5x})^2e^{5x}$$

LESSON 2
Determine $$\displaystyle {dy\over dx}$$ for $$xe^{3y}+y^2=5x^3$$
SOLUTION
• Use the PRODUCT RULE to implicitly differentiate the equation.
$$xe^{3y}+y^2=5x^3$$
$$\displaystyle 1(e^{3y})+x\left(3e^{3y} {dy\over dx}\right)+2y {dy\over dx}=15x^2$$
$$\displaystyle e^{3y}+3xe^{3y} {dy\over dx}+2y {dy\over dx}=15x^2$$
• Group terms with $$\displaystyle {dy\over dx}$$
$$\displaystyle (3xe^{3y}+2y) {dy\over dx}=15x^2-e^{3y}$$
• Solve for $$\displaystyle {dy\over dx}$$
$$\displaystyle {dy\over dx}={15x^2-e^{3y}\over 3xe^3y+2y}$$

LESSON 3
Given that $$y=e^x \sin ⁡x$$, prove that
$$\displaystyle {d^2y\over dx^2}-2 {dy\over dx}+2y=0$$
SOLUTION
• Use the PRODUCT RULE
$$\displaystyle {dy\over dx}=e^x \sin ⁡x+e^x \cos ⁡x$$
$$=e^x(\sin ⁡x+ \cos ⁡x)$$
• Use the PRODUCT RULE again
$$\displaystyle {d^2y\over dx^2}=e^x (\sin ⁡x+\cos ⁡x)+e^x(\cos ⁡x-\sin ⁡x)$$
$$=2e^x \cos ⁡x$$
• Substitute expressions for $$\displaystyle {dy\over dx}$$ and $$\displaystyle {d^2y\over dx^2}$$ into original equation.
$$\displaystyle {d^2 y\over dx^2}-2 {dy\over dx}+2y=2e^x \cos⁡ x-2e^x(\cos ⁡x+\sin ⁡x)+2e^x \sin ⁡x$$
$$=0$$

LESSON 4
Determine $$\displaystyle {dy\over dx}$$ for the equation defined parametrically by
$$x=e^{2t}-1$$ and $$y=\sin^{-1}⁡2t$$.
SOLUTION
$$x=e^{2t}-1$$
$$\displaystyle {dx\over dt}=2e^{2t}$$
$$y=\sin^{-1}⁡2t$$
$$\displaystyle {dy\over dt}={2\over \sqrt {1-(2t)^2}}$$
$$\displaystyle {dy\over dx}={dy\over dt}×{dt\over dx}$$
$$\displaystyle ={2\over \sqrt{1-4t^2}}×{1\over 2e^{2t}}$$
$$\displaystyle ={1\over e^{2t}\sqrt {1-4t^2}}$$