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Lucy (Student)
March 19 2017
Really helpful, good at explaining and patient. Willing to run over if I need it and has really helped my understanding of maths!!
Marion (Parent)
June 10 2017
Marion (Parent)
June 4 2017
Marion (Parent)
May 23 2017
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Questions Larry has answered
Find the derivative of f where f(x)=a^x.
This is a difficult question that you only need to know the result of.However, it's a good exercise to derive it.
Starting with f(x)=ax we can take the natural logarithm of both sides (so we can use one of its properties).
This gives us ln(f(x))=ln(ax), from the natural logarithms properties we know this is equal to ln(f(x))=x*ln(a).
Now using the chain rule we can differentiate both sides,
d(ln(f(x)))/dx= f'(x)/f(x), d(x*ln(a))/dx=ln(a)
so we now have f'(x)/f(x)=ln(a). Recalling that f(x)=ax this gives us the answer,
f'(x)=axln(a).
This is a difficult question that you only need to know the result of.However, it's a good exercise to derive it.
Starting with f(x)=ax we can take the natural logarithm of both sides (so we can use one of its properties).
This gives us ln(f(x))=ln(ax), from the natural logarithms properties we know this is equal to ln(f(x))=x*ln(a).
Now using the chain rule we can differentiate both sides,
d(ln(f(x)))/dx= f'(x)/f(x), d(x*ln(a))/dx=ln(a)
so we now have f'(x)/f(x)=ln(a). Recalling that f(x)=ax this gives us the answer,
f'(x)=axln(a).
Find the inverse of the general 2x2 matrix A= ([a, b],[c, d]) when does this inverse exist?
This is a typical further maths question, doing it correctly is a matter of carrying out a two-step process.
Start by finding the determinant of the matrix,
det(A)=ad-bc
Then swap the entries a d and negate the other entries. After dividing by the determinant the inverse of A is given.
A^-1=1/(ad-bc)([d -b],[-c, a]).
This is a typical further maths question, doing it correctly is a matter of carrying out a two-step process.
Start by finding the determinant of the matrix,
det(A)=ad-bc
Then swap the entries a d and negate the other entries. After dividing by the determinant the inverse of A is given.
A^-1=1/(ad-bc)([d -b],[-c, a]).
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