Using the definition of deriative (f(x+h)-f(x))/h could anyone show me how to calculate the deriative of 4/x^2? I know the answer is -8/x^3 but I don't know the steps to get there.

Guest Oct 29, 2017

#1**+2 **

Using the definition of deriative (f(x+h)-f(x))/h could anyone show me how to calculate the deriative of 4/x^2? I know the answer is -8/x^3 but I don't know the steps to get there.

If you understand the diagram that goes with this, then it is easy to remember because it is totally logical.

Just replace the a with an x in this diagram and see if you can make sense of it.

\(f(x)=\frac{4}{x^2}\\ f(x+h)=\frac{4}{(x+h)^2}\\ \text{The gradient of the secant joining these points }\\ =\frac{rise }{run}\\ =\frac{f(x+h)-f(x)}{x+h-x}\\ =[f(x+h)-f(x)] \div h\\ =[\frac{4}{(x+h)^2}-\frac{4}{x^2}] \div h\\ =[\frac{4}{(x+h)^2}-\frac{4}{x^2}] \div h\\ =\frac{4x^2-4(x+h)^2}{hx^2(x+h)^2}\\ \)

\( =\frac{4x^2-4(x^2+2xh+h^2)}{hx^2(x^2+2xh+h^2)}\\ =\frac{-4(2xh+h^2)}{hx^2(x^2+2xh+h^2)}\\ =\frac{-4h(2x+h)}{hx^2(x^2+2xh+h^2)}\\\)

Now the gradient of the tangent will be the gradient ofthe secant as h tends ot 0

\(\text{gradient of tangent}\\ =\displaystyle\lim_{h\rightarrow 0}\;\;\frac{-4h(2x+h)}{hx^2(x^2+2xh+h^2)}\\ =\displaystyle\lim_{h\rightarrow 0}\;\;\frac{-4(2x+h)}{x^2(x^2+2xh+h^2)}\\ =\dfrac{-8x}{x^4}\\ =\dfrac{-8}{x^3}\)

Melody
Oct 29, 2017