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Parts A and B

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If $y=ax^n$,

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$\\frac{dy}{dx}=anx^{n-1}$ for all rational $n$.

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All we have to do to terms where $x$ is to a power of anything is times the coefficient of $x$ by the original power, and then take one away from the original power.

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Don't be confused if there is no coefficient. The fact the term is there means the coefficient must be $1$, but we don't tend to write it out as, for example $1x$, we just say $x$.

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Part C

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Remember that $\\ln(x)$ differentiates to $\\displaystyle \\frac {1}{x}$. It follows that $a \\ln(x)$ differentiates to $\\displaystyle \\frac {a}{x}$. If the $x$ term is actually, for example, $x+4$ or $2x-1$, then we also need to use the chain rule and multiply our answer by the derivative of the $x$ term. 

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Let's differentiate $3 \\ln(2x-1)$ as an example. Using the fact that $a \\ln(x)$ differentiates to $\\displaystyle \\frac {a}{x}$, we know that the first part of our answer will be

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$\\displaystyle \\frac {3}{2x-1}$

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We then need to find the derivative of $2x-1$ and then multiply the first part of our answer by this. The derivative of $2x-1$ is $2$ so the final answer will be

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$\\displaystyle \\frac {3*2}{2x-1}$ = $\\displaystyle \\frac {6}{2x-1}$

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Part D

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Remember that $\\cos(x)$ differentiates to $-\\sin(x)$. It follows that $a \\cos(x)$ differentiates to $-a \\sin(x)$. Like in the previous example, if the $x$ term is actually something slightly more complicated then we also need to use the chain rule and multiply our answer by the derivative of the $x$ term. 

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Let's differentiate $4 \\cos(3x+1)$ as an example. Using the fact that $a \\cos(x)$ differentiates to $-a \\sin(x)$, we know that the first part of our answer will be

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$\\displaystyle -4 \\sin(3x+1)$

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We then need to find the derivative of $3x+1$ and then multiply the first part of our answer by this. The derivative of $3x+1$ is $3$ so the final answer will be

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$\\displaystyle -4*3 \\sin(3x+1)$ = $\\displaystyle -12 \\sin(3x+1)$

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$y=\\simplify[basic,zeroterm,zerofactor,unitfactor]{{dc[2]}x^{dp}+{ec[2]}x+{fc[2]}}$

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$\\displaystyle \\frac{dy}{dx}=$ [[0]]

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$y=\\simplify[basic,zeroterm,zerofactor,unitfactor]{{dc[3]}x^{dp}+{ec[3]}x+{fc[3]}}$

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$\\displaystyle \\frac{dy}{dx}=$ [[0]]

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$y=\\var{h} \\ln(x+\\var{g})$

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$\\displaystyle \\frac{dy}{dx}=$ [[0]]

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$y=\\var{k} \\cos(\\var{g}x+\\var{l})$

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$\\displaystyle \\frac{dy}{dx}=$ [[0]]

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Differentiate the following functions.

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Differentiating quadratics, natural logs and trig. 

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