Can anyone help me with these two derivatives, I'm a beginner at this?

Use this formula, f(x)=(f(x+h)-f(x))/h, to find the derivatives of,


a) y=(鈭歺-6)


b) y=x/(4-x)Can anyone help me with these two derivatives, I'm a beginner at this?
Hello big Bob, I am little bob. Here's your answers:





f(x)=(f(x+h)-f(x))/h





A) Problem 1. I am assuming y = sqrt(x-6), which means that the 6 is inside the sqrt.


1) The first problem becomes:


lim h--%26gt;0 of (sqrt((x+h)-6)-sqrt(x-6)) / h





This requires a trick to get rid of the h's. The trick is to multiply by the conjugate of the numerator (i.e the conjugate of (a+b) is (a-b)).


The conjugate here is (sqrt((x+h)-6)+sqrt(x-6))





2) Multiply top and bottom by the conjugate:


You get:


( (x+h-6) + (sqrt((x+h)-6)(sqrt(x-6)) - (sqrt((x+h)-6)(sqrt(x-6)) - (x-6) )/


h(sqrt((x+h)-6)+sqrt(x-6))





3) Cancel out the middle 2 terms of numerator...combine...simplify...:


(x+h+6-x+6) / h(sqrt((x+h)-6)+sqrt(x-6))





3.1) Simplify...h/h(sqrt((x+h)-6)+sqrt(x-6)) --%26gt; 1/(sqrt((x+h)-6)+sqrt(x-6))





Now the h's are gone in the messy places (i.e. you won't divide by zero). The main point of all this math is to get rid of h in places where it will force you to divide by zero.


Now as lim h --%26gt; 0, h term is now gone without trouble.





==%26gt; 1/(sqrt((x)-6)+sqrt(x-6)) ==%26gt;(1/2)sqrt(x-6), which is your answer.





==%26gt; If you meant y = sqrt(x)-6, then the answer is: (1/2)sqrt(x), by similar conjugate means (conjugate here is: sqrt(x+h)-6 + sqrt(x)-6).





B) Problem 2:


1) lim h--%26gt;0 of ( (x+h)/(4-x-h) + x/(4-x) ) / h





The key to this problem is to simplify the fraction by multiplying both numerator and denominator by the LCD. You can easily see that the LCD is: (4-x-h)(4-x)





2) If you do so ...combine numerator terms...reduce, here is your result:


( (x+h)(4-x)-x(4-x-h) ) / ( h(4-x)(4-x-h) )





3) Multiply out: (4x-x^2+4h-xh-4x+x^2+xh) / ( h(4-x)(4-x-h) )This reduces to: 4 / (4-x)(4-x-h). Now we can cancel out the bottom h because it will cause us no more trouble as lim h --%26gt; 0.





4) Yielding: 4 / (4-x)(4-x) = 4 / (4-x)^2, which is your answer.





There is no magic bullet for these problems (until you learn the chain rule as well as the product/quotient rules of calculus). However, use intuition and common sense to simplify out h's. It gets easier with experience.Can anyone help me with these two derivatives, I'm a beginner at this?
a) f(x) = 鈭歺 - 6


lim(h鈫?) [ 鈭?x+h) - 6 - (鈭歺 - 6) ] / h =


lim(h鈫?) [ 鈭?x+h) - 6 - 鈭歺 + 6) ] / h =


lim(h鈫?) [ 鈭?x+h) - 鈭歺 ] / h = ................ now rationalize numerator


lim(h鈫?) [ 鈭?x+h) - 鈭歺 ][ 鈭?x+h) + 鈭歺] / h[ 鈭?x+h) + 鈭歺] =


lim(h鈫?) [ x+h - x] / h[ 鈭?x+h) + 鈭歺] =


lim(h鈫?) 1 / [ 鈭?x+h) + 鈭歺] =


1 / (2 鈭歺)





b) f(x) = x / (4 - x)


lim(h鈫?) [ (x+h) / (4 - (x+h)) - x/(4 - x) ] / h =


lim(h鈫?) [ (x+h) / (4 - x- h) - x/(4 - x) ] / h =


lim(h鈫?) [ (x+h)(4-x) - x(4-x-h) ] / [h(4-x)(4-x-h)] =


lim(h鈫?) [ 4x + 4h - x虏 - xh - 4x + x虏 + xh ] / h [16 - 8x + x虏 - 4h - xh] =


lim(h鈫?) [ 4h ] / h[(4 - x)虏 - 4h - xh] =


lim(h鈫?) [ 4 ] / [(4 - x)虏 - 4h - xh] =


4 / (4 - x)虏