How do you differentiate # y = 17(22+x)^((41-x)^30)#?

Question

Emma Aldridge · Accepted Answer

#y'=17(22+x)^((41-x)^30)# Taking the logatithm on both sides we get #ln(y)=ln(17)+(41-x)^30ln(22+x)# differentiating with respect to #x#: #(y')/ y=-30(41-x)^29*ln(22+x)+(41-x)^30*1/(22+x)# so we get #y'=17(22+x)^((41-x)^30)(-30*(41-x)^29*ln(22+x)+(41-x)^30/(22+x))#

Isabella Ackerman · Answer

undefined To differentiate the function $ y = 17(22+x)^{(41-x)^{30}} $ with respect to $ x $, you can use the chain rule along with the power rule for differentiation. The chain rule states that if $ y = f(g(x)) $, then $ \frac{{dy}}{{dx}} = f'(g(x)) \cdot g'(x) $. Applying this rule:

1. Find the derivative of the outer function:
$$ f(u) = u^{30} $$
$$ f'(u) = 30u^{29} $$

2. Find the derivative of the inner function:
$$ g(x) = (41-x) $$
$$ g'(x) = -1 $$

3. Apply the chain rule:
$$ \frac{{dy}}{{dx}} = f'(g(x)) \cdot g'(x) $$
$$ = 30(41-x)^{29} \cdot (-1) $$

Finally, multiply the result with the derivative of the expression inside the parentheses:
$$ = -30(41-x)^{29} \cdot (22+x) $$

So, the derivative of $ y = 17(22+x)^{(41-x)^{30}} $ with respect to $ x $ is:
$$ \frac{{dy}}{{dx}} = -30(41-x)^{29} \cdot (22+x) $$