Solve 1/(h+6)+9

$$\frac { 1 } { ( h + 6 ) + 9 }$$

Differentiate w.r.t. h

$$-\frac{1}{\left(h+15\right)^{2}}$$

Steps Using Chain Rule

Add 6 and 9 to get 15.

$$\frac{\mathrm{d}}{\mathrm{d}h}(\frac{1}{h+15})$$

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).

$$-\left(h^{1}+15\right)^{-1-1}\frac{\mathrm{d}}{\mathrm{d}h}(h^{1}+15)$$

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}.

$$-\left(h^{1}+15\right)^{-2}h^{1-1}$$

Simplify.

$$-h^{0}\left(h^{1}+15\right)^{-2}$$

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

$$-h^{0}\left(h+15\right)^{-2}$$

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

$$-\left(h+15\right)^{-2}$$

Evaluate

$$\frac{1}{h+15}$$

Solution Steps

Add 6 and 9 to get 15.

$$\frac{1}{h+15}$$