Solve (tan(2x+3))^prime

$$( \tan ( 2 x + 3 ) ) ^ { \prime }$$

Evaluate

$$\frac{2\left(\sin(x)\right)^{4}+2\left(\cos(x)\right)^{4}+\left(\sin(3)\sin(2x)\right)^{2}+\left(\cos(3)\sin(2x)\right)^{2}}{\left(\cos(2x+3)\right)^{2}}$$

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).

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

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}.

$$\left(\sec(2x^{1}+3)\right)^{2}\times 2x^{1-1}$$

Simplify.

$$2\left(\sec(2x^{1}+3)\right)^{2}$$

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

$$2\left(\sec(2x+3)\right)^{2}$$

Differentiate w.r.t. x

$$\frac{4\tan(2x+3)\left(22\cos(3)\left(\sin(3)\cos(x)\right)^{2}\left(\sin(x)\right)^{4}-22\cos(3)\left(\sin(3)\sin(x)\right)^{2}\left(\cos(x)\right)^{4}+8\sin(3)\left(\cos(3)\right)^{2}\sin(x)\left(\cos(x)\right)^{5}+8\sin(3)\left(\cos(3)\right)^{2}\cos(x)\left(\sin(x)\right)^{5}+8\left(\cos(3)\right)^{3}\left(\sin(x)\right)^{2}\left(\cos(x)\right)^{4}-8\left(\cos(3)\right)^{3}\left(\cos(x)\right)^{2}\left(\sin(x)\right)^{4}-4\left(\sin(3)\right)^{3}\sin(x)\left(\cos(x)\right)^{5}-4\left(\sin(3)\right)^{3}\cos(x)\left(\sin(x)\right)^{5}+2\cos(3)\left(\sin(3)\right)^{2}\left(\cos(x)\right)^{6}-2\cos(3)\left(\sin(3)\right)^{2}\left(\sin(x)\right)^{6}-4\sin(3)\left(\cos(3)\right)^{2}\left(\sin(2x)\right)^{3}+\left(\sin(3)\sin(2x)\right)^{3}+2\left(\cos(2x+3)\right)^{3}\right)}{\left(\cos(2x+3)\right)^{3}}$$