Differentiate $\frac{\ln(x)}{\cos(x)}$
Please help me with this question.
$$y= \frac{\ln(x)}{\cos(x)}$$
Just starting with calculus. Thank you.
$\endgroup$ 34 Answers
$\begingroup$First compute:
- $f(x) = \ln(x)$ so $f'(x) = 1/x$,
- $g(x) = \cos(x)$ so $g'(x) = -\sin(x)$
then use formula for differentiating quotient $$\left(\frac{f(x)}{g(x)}\right)' = \frac{f'(x)g(x)-g'(x)f(x)}{g^2(x)} = \frac{\frac{\cos(x)}{x}+\sin(x)\ln(x)}{\cos^2(x)} = \frac{\cos(x)+x\ln(x)\sin(x)}{x\cos^2(x)}$$
$\endgroup$ $\begingroup$You have to use the quotient rule: $y' = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2}.$ Let $g(x)=\ln{x}$ and $h(x)=\cos{x}$, then $g'(x)=\frac{1}{x}$ and $h'(x)=-\sin{x}$. After substitution you will have $y'=\frac{cosx+x\sin{x}\ln{x}}{x\cos^2{x}}$.
$\endgroup$ $\begingroup$I'll set up the rule for you, and you finish it, ok? $$y = \frac{(\ln x)' \cdot \cos x - \ln x \cdot (\cos x)'}{(\cos x)^2}$$
$\endgroup$ $\begingroup$As an alternative to the quotient rule, you might also consider rewriting $\frac 1{\cos x}$ as $\sec x$ and using the product rule:
$$\ln (x)\cdot \sec (x)$$
$$\frac{d(f(x)\cdot g(x))}{dx}=f'(x)g(x)+f(x)g'(x)$$
This is a sneaky way of rewriting any quotient rule problem in a product rule form, where $1\over f(x)$ is rewritten as $(f(x))^{-1}$.
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