Solve tan(pi-x)

$$\tan ( \pi - x )$$

Differentiate w.r.t. x

$$-\frac{1}{\left(\cos(x)\right)^{2}}$$

Steps Using Chain Rule

If F is the composition of two differentiable functions f\left(u\right) and u=g\left(x\right), that is, if F\left(x\right)=f\left(g\left(x\right)\right), then the derivative of F is the derivative of f with respect to u times the derivative of g with respect to x, that is, \frac{\mathrm{d}}{\mathrm{d}x}(F)\left(x\right)=\frac{\mathrm{d}}{\mathrm{d}x}(f)\left(g\left(x\right)\right)\frac{\mathrm{d}}{\mathrm{d}x}(g)\left(x\right).

$$\left(\sec(-x^{1}+\pi )\right)^{2}\frac{\mathrm{d}}{\mathrm{d}x}(-x^{1}+\pi )$$

The derivative of a polynomial is the sum of the derivatives of its terms. The derivative of a constant term is 0. The derivative of ax^{n} is nax^{n-1}.

$$\left(\sec(-x^{1}+\pi )\right)^{2}\left(-1\right)x^{1-1}$$

Simplify.

$$-\left(\sec(-x^{1}+\pi )\right)^{2}$$

For any term t, t^{1}=t.

$$-\left(\sec(-x+\pi )\right)^{2}$$

Evaluate

$$-\tan(x)$$