Solve tan(n/4+n/2)

$$\tan ( \frac { n } { 4 } + \frac { n } { 2 } )$$

Differentiate w.r.t. n

$$\frac{3}{4\left(\cos(\frac{3n}{4})\right)^{2}}$$

Steps Using Chain Rule

Combine \frac{n}{4} and \frac{n}{2} to get \frac{3}{4}n.

$$\frac{\mathrm{d}}{\mathrm{d}n}(\tan(\frac{3}{4}n))$$

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).

$$\left(\sec(\frac{3}{4}n^{1})\right)^{2}\frac{\mathrm{d}}{\mathrm{d}n}(\frac{3}{4}n^{1})$$

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}.

$$\left(\sec(\frac{3}{4}n^{1})\right)^{2}\times \frac{3}{4}n^{1-1}$$

Simplify.

$$\frac{3}{4}\left(\sec(\frac{3}{4}n^{1})\right)^{2}$$

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

$$\frac{3}{4}\left(\sec(\frac{3}{4}n)\right)^{2}$$

Evaluate

$$\tan(\frac{3n}{4})$$