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There are various online resources which provide summaries and methods of the different differentiation rules: chain rule, product rule and quotient rule.

To differentiate the functions in this question you have to identify which rules are appropriate. There may be a combination of rules needed.

Example:

For $y=\\ln(2x^3-1)\\cos(4x)+\\frac{1}{3}e^{6x}$

The first term contains both a $\\ln$ function and a $\\cos$ function multiplied together. Differentiating here, therefore, requries the product rule. However, each of these functions contain another function within, so the chain rule will be also required.

The second term is an exponential function with another function in $x$ in the power. This will require the chain rule.

Thus, the process of differentiation is carried out as follows.

$\\frac{dy}{dx}=\\frac{d}{dx}\\left(\\ln(2x^3-1)\\cos(4x)\\right)+\\frac{d}{dx}\\left(\\frac{1}{3}e^{6x}\\right)$

Assign the appropriate subsitutions:

$\\text{Let}\\;t=2x^3-1$

$\\text{Let}\\;u=\\ln(2x^3-1)=\\ln(t)$

$\\text{Let}\\;s=4x$

$\\text{Let}\\;v=\\cos(4x)=\\cos(s)$

$\\text{Let}\\;r=6x$

$\\text{Let}\\;w=\\frac{1}{3}e^{6x}=\\frac{1}{3}e^{r}$

For the first term:

$\\frac{d}{dx}\\left(\\ln(2x^3-1)\\cos(4x)\\right)=u\\frac{dv}{dx}+v\\frac{du}{dx}$

$\\therefore\\;\\;\\;\\frac{d}{dx}\\left(\\ln(2x^3-1)\\cos(4x)\\right)=u\\left(\\frac{dv}{ds}\\times\\frac{ds}{dx}\\right)+v\\left(\\frac{du}{dt}\\times\\frac{dt}{dx}\\right)$

$\\therefore\\;\\;\\;\\frac{d}{dx}\\left(\\ln(2x^3-1)\\cos(4x)\\right)=\\ln(2x^3-1)\\left(-\\sin(4x)\\times4\\right)+\\cos(4x)\\left(\\frac{1}{t}\\times6x^2\\right)$

This is then simplified:

$\\frac{d}{dx}\\left(\\ln(2x^3-1)\\cos(4x)\\right)\\ln(2x^3-1)\\left(-\\sin(4x)\\times4\\right)+\\cos(4x)\\left(\\frac{1}{t}\\times6x^2\\right)=\\frac{6x^{2}cos(4x)}{2x^3-1}-4\\ln(2x^3-1)\\sin(4x)$

For the second term:

$\\frac{d}{dx}\\left(\\frac{1}{3}e^{6x}\\right)=\\frac{dw}{dr}\\times\\frac{dr}{dx}=\\frac{1}{3}e^{6x}\\times6=2e^{6x}$

The final differentiated function is therefore found to be:

$\\left(\\frac{dy}{dx}\\right)=\\frac{6x^{2}cos(4x)}{2x^3-1}-4\\ln(2x^3-1)\\sin(4x)+2e^{6x}$

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$f(x)=\\var{ex[0]}e^x\\sin(\\var{c[0]}x)+x^\\var{ex[1]}$

$f'(x)=$ [[0]]

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$f(x)=\\var{ex[1]}\\sin(\\var{c[1]}x)\\cos(\\var{c[2]}x)$

$f'(x)=$ [[0]]

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$f(x)=$ $\\ln(\\var{ex[2]}x)$$\\var{c[3]}-\\var{c[4]}x$

$f'(x)=$ [[0]]

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$f(x)=\\var{ex[3]}\\sin(\\ln(x))$

$f'(x)=$ [[0]]

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$f(x)=$ $e^{\\var{p}x}$$\\var{c[5]}-x$

$f'(x)=$ [[0]]

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$f(x)=\\ln(\\var{ex[4]}x)\\sin(\\var{ex[5]}x)$

$f'(x)=$ [[0]]

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Find $f'(x)$ of the following, using the appropriate rule.

Do not write out $f'(x)=$.

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extra coeff's added

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Identifying the correct rule to use

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