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Deduzimos as derivadas das inversas das funções exponenciais usando derivação
implícita.
Geometricamente, há uma relação estreita entre os gráficos de \(e^x\) e \(\ln (x)\): eles são reflexões um do outro em relação à reta \(y=x\):
![[Picture]](/rendered/implicitDifferentiation/digInDerivativesOfInverseExponentialFunctions-b62a4848c9b503bc268c0d044a3871a6.svg)
Pode-se suspeitar que é possível usar o fato de que \(\ddx e^x = e^x\) para deduzir a derivada de \(\ln (x)\). Usaremos derivação implícita para explorar essa relação nos cálculos.
A derivada do logaritmo natural \[ \ddx \ln (x) = \frac {1}{x}. \]
Lembre-se de que \[ \ln (x) = y \qquad \text {exatamente quando}\qquad e^y = x\qquad \text {e}\qquad x>\class{xm-given}{0}. \] Logo, \begin{align*} e^y &= x\\ \ddx e^y &= \ddx x &\text {Derive ambos os lados.}\\ e^y \dd [y]{x} &= 1 &\text {Derivação implícita.}\\ \dd [y]{x} &= \frac {1}{e^y} = \class{xm-given}{\frac {1}{x}}&\text {Isole $\dd [y]{x}$.} \end{align*}
Como \(y=\ln (x)\), \(\ddx \ln (x) = \class{xm-given}{\frac {1}{x}}\).
Calcule: \[ \ddx \left (-\ln (\cos (x))\right ) \begin{prompt} = \class{xm-answer}{\cssId{xm-answer-1}{\phantom{\rule{7em}{1.5em}}}} \end{prompt} \]
A partir da derivada do logaritmo natural, podemos deduzir outro fato:
A derivada de um logaritmo qualquer Seja \(b\) um número real positivo. Então \[ \ddx \log _b(x) = \frac {1}{x\ln (b)}. \]
Calcule: \[ \ddx \log _7(x) \begin{prompt} = \class{xm-answer}{\cssId{xm-answer-2}{\phantom{\rule{7em}{1.5em}}}} \end{prompt} \]
Também podemos calcular a derivada de uma função exponencial qualquer.
Calcule: \[ \ddx 7^x \begin{prompt} = \class{xm-answer}{\cssId{xm-answer-3}{\phantom{\rule{7em}{1.5em}}}} \end{prompt} \]