# finding gradient of a multi-variable function

I have to find the gradient of the following function:

$$f(x) = \ln(1 + e^{-ab^Tx})$$

where $f: \mathbb{R}^n \rightarrow \mathbb{R}$, $b \in \mathbb{R}^n$, $a \in \mathbb{R}$ and $x \in \mathbb{R}^n$.

I'm attempting to apply the chain rule here but I'm finding it difficult to apply. The chain rule as I know it goes like:

for $h(x) = f(g(x))$

$$\nabla h(x) = D f(x)^T \nabla g(f(x))$$ where $f :\mathbb{R}^n \rightarrow \mathbb{R}^m$, $g: \mathbb{R}^m \rightarrow \mathbb{R}$ and $h : \mathbb{R}^n \rightarrow \mathbb{R}$

So I set, $g(x) = 1 + e^{-ab^Tx}$ and $f(y) = \ln(y)$. I'm not sure where to go with this now. My guess is that, $$Df(x)^T = \frac{1}{1+e^{-ab^Tx}}$$ and $$\nabla g(f(x)) = -abe^{-ab^Tx}$$ So the final answer should be $$\nabla f(x) = \frac{-abe^{-ab^Tx}}{e^{-ab^Tx}}$$

Is this correct? If wrong, any help to find the right answer would be appreciated. ${{}}$

• You have written $x \in \mathbb R$. I think it should be $x \in \mathbb R^{n}$. – Kavi Rama Murthy Sep 18 '18 at 5:34
• Welcome to math.SE! In addition to the important @KaviRamaMurthy's comment the expression "$f(x) = \ln(1 + e^{-ab^Tx})$" isn't a multivariable function, since only depends on $x$. So in that case you don't have to find a gradient, but the derivative. If not, please correct your grammar, so we can help you better. – manooooh Sep 18 '18 at 6:02
• @KaviRamaMurthy corrected, thanks! – NewComer Sep 18 '18 at 11:29
• @manooooh Since $x \in \mathbb{R}^n$ I thought it belongs to multi variable calculus. Isn't it? If not, what do you call these calculations? – NewComer Sep 18 '18 at 11:30
• @manooooh Surely $b^T x \in \mathbb{R}^n \times \mathbb{R}^n$ – Kevin Sep 18 '18 at 11:33

$$D (f \circ g)(x) = Df(g(x)) \circ Dg(x)$$ (See the Wikipedia page on the chain rule. $$Df(x)$$ means the Jacobian matrix of $$f$$ at the point $$x$$)
I would decompose your function as $$F(x) = f(g(x))$$ where $$g(x) = b^T x$$ and $$f(y) = \ln(1+e^{ay})$$. Since $$g(x) = b_1x_1 + \cdots b_n x_n$$, then $$Dg(x) = \nabla g(x) = b$$. Also, $$Df(y) = \frac{df}{dy} = \frac{ae^{ay}}{1+e^{ay}}$$. So I think you get
$$\nabla F(x) = \frac{ae^{ab^Tx}}{1+e^{ab^Tx}} \, b$$