Calculus Examples

$J_{j (θ)} = - y^{(j)} \log (σ) (x^{(j)} \cdot θ) - (1 - y^{(j)}) \log (1 - σ (x^{(j)} \cdot θ))$

Step 1

This derivative could not be completed using the chain rule. Mathway will use another method.

Step 2

Differentiate using the Sum Rule.

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Step 2.1

Remove parentheses.

$\frac{d}{d y} [- y^{(j)} \log (σ) (x^{j} \cdot θ) - (1 - y^{(j)}) \log (1 - σ x^{j} θ)]$

Step 2.2

Reorder factors in $\log (1 - σ x^{j} θ)$ .

$\frac{d}{d y} [- y^{(j)} \log (σ) (x^{j} \cdot θ) - (1 - y^{(j)}) \log (1 - σ θ x^{j})]$

Step 2.3

By the Sum Rule, the derivative of $- y^{(j)} \log (σ) (x^{j} \cdot θ) - (1 - y^{(j)}) \log (1 - σ θ x^{j})$ with respect to $y$ is $\frac{d}{d y} [- y^{(j)} \log (σ) (x^{j} \cdot θ)] + \frac{d}{d y} [- (1 - y^{(j)}) \log (1 - σ θ x^{j})]$ .

$\frac{d}{d y} [- y^{(j)} \log (σ) (x^{j} \cdot θ)] + \frac{d}{d y} [- (1 - y^{(j)}) \log (1 - σ θ x^{j})]$

Step 3

Evaluate $\frac{d}{d y} [- y^{(j)} \log (σ) (x^{j} \cdot θ)]$ .

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Step 3.1

Since $- \log (σ) x^{j} θ$ is constant with respect to $y$ , the derivative of $- y^{j} \log (σ) (x^{j} θ)$ with respect to $y$ is $- \log (σ) x^{j} θ \frac{d}{d y} [y^{j}]$ .

$- \log (σ) x^{j} θ \frac{d}{d y} [y^{j}] + \frac{d}{d y} [- (1 - y^{(j)}) \log (1 - σ θ x^{j})]$

Step 3.2

Differentiate using the Power Rule which states that $\frac{d}{d y} [y^{n}]$ is $n y^{n - 1}$ where $n = j$ .

$- \log (σ) x^{j} θ (j y^{j - 1}) + \frac{d}{d y} [- (1 - y^{(j)}) \log (1 - σ θ x^{j})]$

Step 4

Evaluate $\frac{d}{d y} [- (1 - y^{(j)}) \log (1 - σ θ x^{j})]$ .

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Step 4.1

Since $- \log (1 - σ θ x^{j})$ is constant with respect to $y$ , the derivative of $- (1 - y^{j}) \log (1 - σ θ x^{j})$ with respect to $y$ is $- \log (1 - σ θ x^{j}) \frac{d}{d y} [1 - y^{j}]$ .

$- \log (σ) x^{j} θ (j y^{j - 1}) - \log (1 - σ θ x^{j}) \frac{d}{d y} [1 - y^{j}]$

Step 4.2

By the Sum Rule, the derivative of $1 - y^{j}$ with respect to $y$ is $\frac{d}{d y} [1] + \frac{d}{d y} [- y^{j}]$ .

$- \log (σ) x^{j} θ (j y^{j - 1}) - \log (1 - σ θ x^{j}) (\frac{d}{d y} [1] + \frac{d}{d y} [- y^{j}])$

Step 4.3

Since $1$ is constant with respect to $y$ , the derivative of $1$ with respect to $y$ is $0$ .

$- \log (σ) x^{j} θ (j y^{j - 1}) - \log (1 - σ θ x^{j}) (0 + \frac{d}{d y} [- y^{j}])$

Step 4.4

Since $- 1$ is constant with respect to $y$ , the derivative of $- y^{j}$ with respect to $y$ is $- \frac{d}{d y} [y^{j}]$ .

$- \log (σ) x^{j} θ (j y^{j - 1}) - \log (1 - σ θ x^{j}) (0 - \frac{d}{d y} [y^{j}])$

Step 4.5

Differentiate using the Power Rule which states that $\frac{d}{d y} [y^{n}]$ is $n y^{n - 1}$ where $n = j$ .

$- \log (σ) x^{j} θ (j y^{j - 1}) - \log (1 - σ θ x^{j}) (0 - (j y^{j - 1}))$

Step 4.6

Subtract $j y^{j - 1}$ from $0$ .

$- \log (σ) x^{j} θ (j y^{j - 1}) - \log (1 - σ θ x^{j}) (- j y^{j - 1})$

Step 4.7

Multiply $- 1$ by $- 1$ .

$- \log (σ) x^{j} θ (j y^{j - 1}) + 1 \log (1 - σ θ x^{j}) (j y^{j - 1})$

Step 4.8

Multiply $\log (1 - σ θ x^{j})$ by $1$ .

$- \log (σ) x^{j} θ (j y^{j - 1}) + \log (1 - σ θ x^{j}) (j y^{j - 1})$

Step 5

Simplify.

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Step 5.1

Reorder terms.

$- y^{j - 1} x^{j} j θ \log (σ) + y^{j - 1} j \log (1 - σ θ x^{j})$

Step 5.2

Reorder factors in $- y^{j - 1} x^{j} j θ \log (σ) + y^{j - 1} j \log (1 - σ θ x^{j})$ .

$- j θ y^{j - 1} x^{j} \log (σ) + j y^{j - 1} \log (1 - σ θ x^{j})$