What is the equation of the line tangent to the curve #y=(x^2-35)^7# at x=6?

Question

Evelyn Agard · Accepted Answer

The tangent line in a point to a function, knowing its ascissa is:

#y-f(x_0)=f'(x_0)(x-x_0)#.

So:

#f(x_0)=f(6)=(36-35)^7=1#

and

#f'(x)=7(x^2-36)^6*2xrArr#

#f'(6)=7(36-35)^6*2*6=7*1*12=84#.

The tangent line is:

#y-1=84(x-6)rArry=84x-503#.

Christopher Allers · Answer

Emma Aldridge · Answer

undefined To find the equation of the line tangent to the curve y=(x^2-35)^7 at x=6, we need to find the derivative of the function y=(x^2-35)^7 and evaluate it at x=6. The derivative of y=(x^2-35)^7 can be found using the chain rule, which states that the derivative of (f(g(x))) is equal to f'(g(x)) * g'(x). Applying the chain rule, we get:

dy/dx = 7(x^2-35)^6 * 2x

Evaluating this derivative at x=6, we have:

dy/dx = 7(6^2-35)^6 * 2(6)

Simplifying this expression, we get:

dy/dx = 7(36-35)^6 * 12

dy/dx = 7(1)^6 * 12

dy/dx = 7 * 12

dy/dx = 84

Therefore, the slope of the tangent line at x=6 is 84. To find the equation of the line, we can use the point-slope form of a line, which is y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope. Since the point of tangency is (6, (6^2-35)^7), we can substitute these values into the equation:

y - (6^2-35)^7 = 84(x - 6)

Simplifying this equation will give us the equation of the line tangent to the curve y=(x^2-35)^7 at x=6.