Correct antiderivative of $(7x+9)^{1/2}$

Question

I have $f(x) = (7x+9)^\frac{1}{2}$

Now, as per the power rule, the antiderivative (indefinite integral) for the above function will be $f(x) = \frac{2}{3}\cdot(7x+9)^{\frac{3}{2}}$ However, if I differentiate $f(x) = f^{\prime}(x)$ then I get $f(x) = 7\times(7x+9)^{\frac{1}{2}}$

I am getting confused about how there are two $f(x)$ for the same $F(x)$.

As per my understanding $f(x)$ -> antiderivative -> $F(x)$ -> differentiate/$F^{\prime}(x)$ -> $f(x)$

How is differentiating $F(x)$ returns two separate functions? Can someone please clarify?

NOTE: I have tried u-substitution where $u$ was $7x+9$.

Answer 1

The problem here is that you are assuming that since the antiderivative of $x^{1/2}$ is $\frac23x^{3/2}$ (which is indeed correct) also the antiderivative of $g(x)^{1/2}$ will be $\frac23g(x)^{3/2}$.

This is indeed wrong as $$\left(\frac23g(x)^{3/2}\right)'=g(x)^{1/2}\cdot g'(x)$$ so the inner derivative is the one causing the issue.

In general there is no "chain rule" for integration (otherwise integration would be easy) besides the one for affine functions, which states that $$\int f(ax+b)=\frac1{a}F(ax+b)$$

Answer 2

Assuming you mean $\int\sqrt{7x+9}dx$:

Substitute $u=7x+9$, $du=\frac{dx}{7}$

$$\int\sqrt{7x+9}dx=\frac{1}{7}\int\sqrt{u}du=\frac{2}{3(7)}u^{\frac{3}{2}}+C=\frac{2}{21}(7x+9)^{\frac{3}{2}}+C$$

Then, $f'(x)=\frac{2}{21}(\frac{3}{2})(7x+9)^\frac{1}{2}\times\frac{d}{dx}(7x+9)=\frac{1}{7}(7x+9)^\frac{1}{2}(7)=(7x+9)^{1/2}$

I'm guessing you forgot to multiply by $\frac{d}{dx}(7x+9)$ while using the chain rule.