Solve cos60y/16

$$\cos 60 \frac { y } { 16 }$$

Differentiate w.r.t. y

$$-\frac{15\sin(\frac{15y}{4})}{4}$$

Steps Using Chain Rule

Express 60\times \frac{y}{16} as a single fraction.

$$\frac{\mathrm{d}}{\mathrm{d}y}(\cos(\frac{60y}{16}))$$

Divide 60y by 16 to get \frac{15}{4}y.

$$\frac{\mathrm{d}}{\mathrm{d}y}(\cos(\frac{15}{4}y))$$

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).

$$\left(-\sin(\frac{15}{4}y^{1})\right)\frac{\mathrm{d}}{\mathrm{d}y}(\frac{15}{4}y^{1})$$

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}.

$$\left(-\sin(\frac{15}{4}y^{1})\right)\times \frac{15}{4}y^{1-1}$$

Simplify.

$$-\frac{15}{4}\sin(\frac{15}{4}y^{1})$$

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

$$-\frac{15}{4}\sin(\frac{15}{4}y)$$

Evaluate

$$\cos(\frac{15y}{4})$$