Solve f(x)=1/4x+1

$$f ( x ) = \frac { 1 } { 4 x + 1 }$$

Differentiate w.r.t. x

$$-\frac{4}{\left(4x+1\right)^{2}}$$

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).

$$-\left(4x^{1}+1\right)^{-1-1}\frac{\mathrm{d}}{\mathrm{d}x}(4x^{1}+1)$$

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}.

$$-\left(4x^{1}+1\right)^{-2}\times 4x^{1-1}$$

Simplify.

$$-4x^{0}\left(4x^{1}+1\right)^{-2}$$

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

$$-4x^{0}\left(4x+1\right)^{-2}$$

For any term t except 0, t^{0}=1.

$$-4\left(4x+1\right)^{-2}$$

Evaluate

$$\frac{1}{4x+1}$$