- 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
\(x=e^{2t}-1\)
\(\displaystyle {dx\over dt}=2e^{2t}\)
\(y=\sin^{-1}⁡2t\)
\(\displaystyle {dy\over dt}={2\over \sqrt {1-(2t)^2}}\)
\(\displaystyle {dy\over dx}={dy\over dt}×{dt\over dx}\)
\(\displaystyle ={2\over \sqrt{1-4t^2}}×{1\over 2e^{2t}}\)
\(\displaystyle ={1\over e^{2t}\sqrt {1-4t^2}}\)