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$7^{x^2-x}$

I know there is a formula ($(a^u)' = u' \cdot a^u \cdot \ln a$) for this but I wanted to understand the logic behind that formula so I tried using the chain rule to solve this:

$$(7^{x^2-x})' = (x^2-x)'\cdot7^{x^2-x-1}\cdot 7' = (2x-1)\cdot7^{x^2-x-1}\cdot0 = 0$$

Then I tried replacing $7^{x^2-x}$ with $(e^x)^{\ln7\cdot(x-1)}$:

$$((e^x)^{\ln 7 (x-1)})' = (\ln 7 (x-1))' \cdot (e^x)^{\ln 7 (x-1)-1} \cdot (e^x)' = (\ln7) \cdot (e^x)^{\ln 7 (x-1)-1} \cdot e^x = (\ln7) \cdot (e^x)^{\ln 7 (x-1)} = (\ln7) \cdot 7^{x(x-1)} =(\ln7) \cdot 7^{x(x-1)}$$

But the right answer is $(\ln7) \cdot 7^{x(x-1)} \cdot (2x-1)$. What did I do wrong?

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    $e^{xln(7x-7)} $ this is wrong, the right is : $e^{(x^2-x)ln7}$ – 2017-01-13

2 Answers 2

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You can go straight to the chain rule.

Call $f(x)=7^x$ and $g(x)=x^2-x$, then

$$g'(x)=2x-1 \quad \text{and} \quad f'(x)=7^x(\ln 7)$$

so,

$$[f(g(x))]'=g'(x)f'(g(x))=(2x-1)7^{g(x)}(\ln7)=(2x-1)7^{x(x-1)}(\ln7)$$

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    @SilenceOnTheWire: is it clear? – 2017-01-21
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As you have written $$(a^u)' = u' \cdot a^u \cdot \ln a$$ so :$$(a^{ u })'=u'\cdot a^{ u }\cdot \ln { a } ={ \left( 7^{ x^{ 2 }-x } \right) }^{ \prime }={ \left( { x }^{ 2 }-x \right) }^{ \prime }7^{ x^{ 2 }-x }\ln { 7 } =\left( 2x-1 \right) { 7 }^{ { x }^{ 2 }-x }\ln { 7 } $$