How is the derivative of $(5x-2)^3$ equal to $15(5x-2)^2$ and not $3(5x-2)^2$. According to $\frac{df}{dx} = nx^{n-1}$, it has to be $3(5x-2)^2$ right. Please explain.
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$\begingroup$Hint: By the chain rule we have that $$\frac{d}{dx}f(g(x))=f'(g(x))g'(x).$$
In your question the answer would be $3(5x-2)^2$ if you were taking the derivative with respect to $5x-2$, but you are not, instead you are taking the derivative with respect to $x$.
$\endgroup$ $\begingroup$You simply forgot to remember using the chain rule: \begin{align} \frac{d}{dx}(5x-2)^3 = 3(5x-2)^2 \frac{d}{dx}(5x-2) =3(5x-2)^2 \cdot 5 = \boxed{15(5x-2)^2} \end{align}
$\endgroup$ $\begingroup$No, $$\frac{d}{dx} x^n = n x^{n - 1}$$ only applies to pure powers of $x$. $5x - 2$ is not a pure power of $x$, so this does not apply. The chain rule is necessary here; the derivative of the inner function ($5x - 2$) is $5$ here, accounting for the additional factor.
$\endgroup$ $\begingroup$We need to use the rule:
$\frac{d}{dx}(f(x)^n)=n f'(x)f(x)^{n-1}$, where $f(x)=5x-2$, and $n=3$, so $f'(x)=5$
Combining all of this we have:
$\frac{d}{dx}((5x-2)^3)=(3)(5)(5x-2^{3-1})=15(5x-2)^2$
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