Currently unavailable: for new students
Degree: History (Bachelors) - Birmingham University
Hi, I'm Sophie and I'm a first year History student. I have always had a passion for history as looking at the past often gives a greater degree of understading of our present world and hope to instill this passion in my students too!
My expertise are in writing essays and coursework, including forming arguments and helping with grammar, as this is where I excelled in A Level History and English Literature. My subject expertise include Tsarist Russia, the Russian Revolution and the Tudors.
I am patient and able to offer well thought out explanations. I try my best to make all topics fun! I have experience helping and teaching people of all ages as I ran orchestras and sports clubs in my spare time at secondary school and assisted a GCSE Maths class during my A Levels. This also helped me understand differences in the way people learn.
The focus of our sessions will be you! You will guide your own learning and what you would like us to cover in each session. Whether you need help understanding a topic, or would like me to read over some coursework, or help form an argument for an essay, I am happy to help!
|English Literature||GCSE||£18 /hr|
|-Personal Statements-||Mentoring||£20 /hr|
|Before 12pm||12pm - 5pm||After 5pm|
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This is a typical gradient question for A2 papers, as it requires use of the Product Rule and Chain Rule, as well as knowledge of e^x.
Firstly, we can see that we will need to use the Product Rule as e^2-x and ln(3x-2) are being multiplied together. As the Product Rule is dy/dx= u.dv/dx + v.du/dx, we will also need to find out the differentials of e^2-x and ln(3x-2), where we let e^2-x = u and ln(3x-2) = v.
u=e^2-x This is 2 a combination of operations, meaning we will need the Chain Rule to obtain du/dx.If we let 2-x=t, then u=e^t. Differentiating 2-x mans that dt/dx= -1 and differentiating e^t means that du/dt remains e^t. The Chain Rule is du/dx=dt/dx*du/dt, meaning that du/dx=-1*e^t, and substituting t back into the eqution leaves us with du/dx=-e^2-x.
v=ln(3x-2) Again, we have 2 operations meaning the Chain Rule will be used to obtain dv/dx. If we let 3x-2=m, then v=ln(m). Differentiating 3x-2 means dm/dx=3, and differentiating ln(m) means dv/dm=1/m. Again, we will need to use the Chain Rule, dv/dx=dm/dx*dv/dm, meaning dv/dx=3*1/m. Substituting our m back into this equation leaves us with dv/dx=3/3x-2.
As we have now obtained our du/dx and our dv/dx, we cn use the Product Rule to find the gradient of our original function. The Product Rule is dy/dx=u.dv/dx+v.du/dx, meaning dy/dx=(e^2-x)(3/3x-2)+(ln(3x-2))(-e^2-x).
As we have an x value given to us in the question, we do not need to simplify this eqution and can cimply substitute our x value of 2 into this equation to find the gradient of the curve at that point. With our substituted x value of 2, dy/dx=(e^0)(3/4)+(ln(4))(-e^0)=(1)(3/4)+(ln(4))(-1). Therefore, our final answer is dy/dx=3/4-ln(4) at on the point on the curve where x=2. Using our Laws of Logarithms, this can also be written as dy/dx=3/4+ln(4)^-1=3/4+ln(1/4), and either answer would be ccepted in the exam.see more